A. Apple

按题意模拟即可。

  1. #include<stdio.h>
  2. #include<iostream>
  3. #include<string.h>
  4. #include<string>
  5. #include<ctype.h>
  6. #include<math.h>
  7. #include<set>
  8. #include<map>
  9. #include<vector>
  10. #include<queue>
  11. #include<bitset>
  12. #include<algorithm>
  13. #include<time.h>
  14. using namespace std;
  15. void fre() { freopen("c://test//input.in", "r", stdin); freopen("c://test//output.out", "w", stdout); }
  16. #define MS(x, y) memset(x, y, sizeof(x))
  17. #define ls o<<1
  18. #define rs o<<1|1
  19. typedef long long LL;
  20. typedef unsigned long long UL;
  21. typedef unsigned int UI;
  22. template <class T1, class T2>inline void gmax(T1 &a, T2 b) { if (b > a)a = b; }
  23. template <class T1, class T2>inline void gmin(T1 &a, T2 b) { if (b < a)a = b; }
  24. const int N = 105, M = 0, Z = 1e9 + 7, inf = 0x3f3f3f3f;
  25. template <class T1, class T2>inline void gadd(T1 &a, T2 b) { a = (a + b) % Z; }
  26. int casenum, casei;
  27. int n, m;
  28. vector<int>a[N];
  29.  
  30. int ind1 = 0, ind1num = 0;
  31. int ind2num = 0;
  32. int ind3 = 0, ind3num = 0;
  33. bool vis[N];
  34. int ind[N];
  35.  
  36. int cnt;
  37. bool dfs1(int x, int fa)
  38. {
  39.     ++cnt;
  40.     vis[x] = 1;
  41.     for(auto y : a[x])if(y != fa)
  42.     {
  43.         if(!vis[y])
  44.         {
  45.             if(!dfs1(y, x))return 0;
  46.         }
  47.         else if(y != ind3)return 0;
  48.     }
  49.     return 1;
  50. }
  51.  
  52. void dfs3(int x, int fa)
  53. {
  54.     ++cnt;
  55.     vis[x] = 1;
  56.     for(auto y : a[x])if(y != fa)
  57.     {
  58.         if(!vis[y])
  59.         {
  60.             dfs3(y, x);
  61.         }
  62.     }
  63. }
  64.  
  65. bool solve()
  66. {
  67.     if(ind1num != 1)return 0;
  68.     if(ind2num != n - 2)return 0;
  69.     if(ind3num != 1)return 0;
  70.  
  71.     cnt = 0;
  72.     vis[ind3] = true;
  73.     if(!dfs1(ind1, 0))return 0;
  74.     dfs3(ind3, 0);
  75.     if(cnt != n)return 0;
  76.  
  77.     return 1;
  78. }
  79. int main()
  80. {
  81.     freopen("input.txt", "r", stdin);
  82.     freopen("output.txt", "w", stdout);
  83. 	while(~scanf("%d%d", &n, &m))
  84. 	{
  85.         for(int i = 1; i <= n; ++i)
  86.         {
  87.             a[i].clear();
  88.             ind[i] = 0;
  89.             vis[i] = 0;
  90.         }
  91.         for(int i = 1; i <= m; ++i)
  92.         {
  93.             int x, y; scanf("%d%d", &x, &y);
  94.             a[x].push_back(y);
  95.             a[y].push_back(x);
  96.             ++ind[x];
  97.             ++ind[y];
  98.         }
  99.         ind1 = 0, ind1num = 0;
  100.         ind2num = 0;
  101.         ind3 = 0, ind3num = 0;
  102.         for(int i = 1; i <= n; ++i)
  103.         {
  104.             if(ind[i] == 1)
  105.             {
  106.                 ++ind1num;
  107.                 ind1 = i;
  108.             }
  109.             else if(ind[i] == 2)
  110.             {
  111.                 ++ind2num;
  112.             }
  113.             else if(ind[i] == 3)
  114.             {
  115.                 ++ind3num;
  116.                 ind3 = i;
  117.             }
  118.         }
  119.         puts(solve() ? "Yes" : "No");
  120. 	}
  121.  
  122. 	return 0;
  123. }
  124. /*
  125. 【trick&&吐槽】
  126.  
  127. 【题意】
  128.  
  129. 【分析】
  130.  
  131. 【时间复杂度&&优化】
  132. 9 9
  133. 1 2
  134. 2 3
  135. 3 5
  136. 5 6
  137. 6 4
  138. 4 7
  139. 1 7
  140. 7 9
  141. 8 9
  142.  
  143. 5 5
  144. 1 2
  145. 2 3
  146. 1 3
  147. 1 4
  148. 1 5
  149.  
  150. 4 4
  151. 1 2
  152. 2 3
  153. 1 3
  154. 1 4
  155.  
  156. 5 4
  157. 1 2
  158. 2 3
  159. 1 3
  160. 1 4
  161.  
  162. */

　　

B. Bar charts

关于序列的前缀和建立差分约束系统，SPFA判断是否存在负环。

  1. #include<cstdio>
  2. #include<cstdlib>
  3. using namespace std;
  4. const int N=12222,M=1000000,E=10000000,inf=~0U>>1;
  5. int n,i,g[N],v[M],w[M],nxt[M],ed,d[N];
  6. int x,h,t,q[E];
  7. int vis[N],in[N];
  8. inline void add(int x,int y,int z){v[++ed]=y;w[ed]=z;nxt[ed]=g[x];g[x]=ed;}
  9. inline void make(int x,int y,int z){
  10.     //x-y=z
  11.     add(y,x,z);
  12.     add(x,y,-z);
  13. }
  14. void gao(){
  15.     int k,b,s;
  16.     scanf("%d%d%d",&k,&b,&s);
  17.     make(b-1,0,0);
  18.     for(int i=b+k*s;i<=n;i++)make(i,i-1,0);
  19.     for(int i=1;i<=k;i++){
  20.         int l=b+(i-1)*s,r=b+i*s;
  21.         int o;
  22.         scanf("%d",&o);
  23.         //s[r-1]-s[l-1]
  24.         make(r-1,l-1,o);
  25.     }
  26. }
  27. void NO(){
  28.     puts("No");
  29.     exit(0);
  30. }
  31. inline void ext(int x,int y){
  32.     if(y<0)NO();
  33.     if(y>=d[x])return;
  34.     d[x]=y;
  35.     if(!in[x]){
  36.         in[x]=1;
  37.         vis[x]++;
  38.         if(vis[x]>=n)NO();
  39.         q[++t]=x;
  40.         if(t>=E-10)NO();
  41.     }
  42. }
  43. int main(){
  44.     freopen("input.txt","r",stdin);
  45.     freopen("output.txt","w",stdout);
  46.     n=11111;
  47.     for(i=0;i<n;i++)add(i+1,i,0);
  48.     gao();
  49.     gao();
  50.     for(i=0;i<=n;i++){
  51.         d[i]=inf;
  52.     }
  53.     h=1,t=0;
  54.     ext(0,0);
  55.     while(h<=t){
  56.         x=q[h++];
  57.         for(i=g[x];i;i=nxt[i])ext(v[i],d[x]+w[i]);
  58.         in[x]=0;
  59.     }
  60.     puts("Yes");
  61. }

　　

C. Construction sets

二分答案，二进制拆分背包+bitset检验。

  1. #include<stdio.h>
  2. #include<iostream>
  3. #include<string.h>
  4. #include<string>
  5. #include<ctype.h>
  6. #include<math.h>
  7. #include<set>
  8. #include<map>
  9. #include<vector>
  10. #include<queue>
  11. #include<bitset>
  12. #include<algorithm>
  13. #include<time.h>
  14. using namespace std;
  15. void fre() { freopen("c://test//input.in", "r", stdin); freopen("c://test//output.out", "w", stdout); }
  16. #define MS(x, y) memset(x, y, sizeof(x))
  17. #define ls o<<1
  18. #define rs o<<1|1
  19. typedef long long LL;
  20. typedef unsigned long long UL;
  21. typedef unsigned int UI;
  22. template <class T1, class T2>inline void gmax(T1 &a, T2 b) { if (b > a)a = b; }
  23. template <class T1, class T2>inline void gmin(T1 &a, T2 b) { if (b < a)a = b; }
  24. const int N = 55, M = 0, Z = 1e9 + 7, inf = 0x3f3f3f3f;
  25. template <class T1, class T2>inline void gadd(T1 &a, T2 b) { a = (a + b) % Z; }
  26. int casenum, casei;
  27. int n, mn, mx;
  28. int m[N], c[N];
  29. bitset<10005>f;
  30. bool check(int K)
  31. {
  32.     f.reset(); f[0] = 1;
  33.     for(int i = 1; i <= n; ++i)
  34.     {
  35.         int tot = c[i] / K;
  36.         int g = 1;
  37.         LL v = m[i];
  38.         while(tot)
  39.         {
  40.             if(v > mx)break;
  41.             f |= f << v;
  42.             tot -= g;
  43.             g = min(g * 2, tot);
  44.             v = (LL)m[i] * g;
  45.         }
  46.     }
  47.     for(int j = mn; j <= mx; ++j)
  48.     {
  49.         if(f[j])return 1;
  50.     }
  51.     return 0;
  52. }
  53. int main()
  54. {
  55.     freopen("input.txt", "r", stdin);
  56.     freopen("output.txt", "w", stdout);
  57. 	while(~scanf("%d%d%d", &n, &mn, &mx))
  58. 	{
  59.         for(int i = 1; i <= n; ++i)
  60.         {
  61.             scanf("%d%d", &m[i], &c[i]);
  62.         }
  63.         int l = 0;
  64.         int r = 1e6;
  65.         while(l < r)
  66.         {
  67.             int mid = (l + r + 1) >> 1;
  68.             if(check(mid))
  69.             {
  70.                 l = mid;
  71.             }
  72.             else
  73.             {
  74.                 r = mid - 1;
  75.             }
  76.         }
  77.         printf("%d\n", l);
  78. 	}
  79.  
  80. 	return 0;
  81. }
  82. /*
  83. 【trick&&吐槽】
  84.  
  85. 【题意】
  86.  
  87. 【分析】
  88.  
  89. 【时间复杂度&&优化】
  90. 3 12 13
  91. 3 8
  92. 4 6
  93. 7 9
  94.  
  95. */

　　

D. Dinner party

$f[i][j]$表示面积和为$i$的矩形，周长和为$j$是否有可能，bitset加速。

  1. #include<cstdio>
  2. #include<bitset>
  3. using namespace std;
  4. const int N=1010,M=2010;
  5. int T,n,m,i,j,x,y,cnt;
  6. bitset<M>f[N];
  7. void solve(){
  8.     scanf("%d%d",&n,&m);
  9.     if(m%2){
  10.         puts("No");
  11.         return;
  12.     }
  13.     m/=2;
  14.     if(f[n][m]==0){
  15.         puts("No");
  16.         return;
  17.     }
  18.     puts("Yes");
  19.     int cnt=0;
  20.     static int q[100000][2];
  21.     while(n){
  22.         int X=0,Y=0;
  23.         for(x=1;x<=n;x++){
  24.             for(y=1;y*x<=n;y++)if(x+y<=m)if(f[n-x*y][m-x-y]){
  25.                 X=x,Y=y;
  26.                 break;
  27.             }
  28.             if(X)break;
  29.         }
  30.         q[++cnt][0]=X;
  31.         q[cnt][1]=Y;
  32.         n-=X*Y;
  33.         m-=X+Y;
  34.     }
  35.     printf("%d\n",cnt);
  36.     for(int i=1;i<=cnt;i++)printf("%d %d\n",q[i][0],q[i][1]);
  37. }
  38. int main(){
  39.     freopen("input.txt","r",stdin);
  40.     freopen("output.txt","w",stdout);
  41.     //scanf("%d%d",&n,&m);
  42.     n=1000;
  43.     m=2000;
  44.     f[0][0]=1;
  45.     for(i=0;i<=n;i++)
  46.         for(x=1;x+i<=n;x++)
  47.             for(y=1;x*y+i<=n;y++)
  48.                 f[i+x*y]|=f[i]<<(x+y);
  49.     scanf("%d",&T);
  50.     while(T--)solve();
  51. }

　　

E. Expression on dice

若要构造$=$，则在两侧构造$0$即可，否则随意构造。

  1. //
  2. //  main.cpp
  3. //  opencup 10378 E
  4. //
  5. //  Created by luras on 2017/9/14.
  6. //  Copyright © 2017年 luras. All rights reserved.
  7. //
  8.  
  9. #define ms(x, y) memset(x, y, sizeof(x))
  10. #define mc(x, y) memcpy(x, y, sizeof(x))
  11. #define lson o << 1, l, mid
  12. #define rson o << 1 | 1, mid + 1, r
  13. #define ls o << 1
  14. #define rs o << 1 | 1
  15. #include<stdio.h>
  16. #include<string.h>
  17. #include<math.h>
  18. #include<queue>
  19. #include<map>
  20. #include<stack>
  21. #include<time.h>
  22. #include<vector>
  23. #include<list>
  24. #include<set>
  25. #include<iostream>
  26. #include<stdlib.h>
  27. #include<string>
  28. #include<algorithm>
  29. #pragma comment(linker,"/STACK:102400000,102400000")
  30. template <class T> inline void gmax(T &a, T b){if(b > a) a = b;}
  31. template <class T> inline void gmin(T &a, T b){if(b < a) a = b;}
  32. using namespace std;
  33. const int N = 1e6 + 10, M = 2e6 + 10, Z = 1e9 + 7, maxint = 2147483647, ms1 = 16843009, ms31 = 522133279, ms63 = 1061109567, ms127 = 2139062143;
  34. const double PI = acos(-1.0), eps = 1e-8;
  35. typedef long long LL;
  36. void fre()
  37. {
  38.     freopen("/Users/luras/Desktop/in.txt", "r", stdin);
  39.     freopen("/Users/luras/Desktop/out.txt", "w", stdout);
  40. }
  41. const int INF = 1e9;
  42. int casenum, casei;
  43.  
  44. int priv[N];
  45. double calc(double a, double b, char op)
  46. {
  47.     if(op == '+') return a + b;
  48.     if(op == '-') return a - b;
  49.     if(op == '*') return a * b;
  50.     if(op == '/') return a / b;
  51.     return 0;
  52. }
  53.  
  54. double calculate(string str)
  55. {
  56.     stack<double> num;
  57.     stack<char> oper;
  58.     priv['+'] = priv['-'] = 3;
  59.     priv['*'] = priv['/'] = 2;
  60.     priv['('] = 10;
  61.     double x, y, t = 0;
  62.     int i; char last = 0;
  63.     for(i = 0; i < str.length(); i ++){
  64.         if(isdigit(str[i])){
  65.             num.push(atof(str.c_str() + i));
  66.             for(; i + 1 < str.length() && isdigit(str[i + 1]); i ++);
  67.             if(i + 1 < str.length() && str[i + 1] == '.')
  68.                 for(i ++; i + 1 < str.length() && isdigit(str[i + 1]); i ++);
  69.         }
  70.         else if(str[i] == '('){
  71.             oper.push(str[i]);
  72.         }
  73.         else if(str[i] == ')'){
  74.             while(oper.top() != '('){
  75.                 y = num.top(); num.pop();
  76.                 x = num.top(); num.pop();
  77.                 char op = oper.top();
  78.                 oper.pop();
  79.                 num.push(calc(x, y, op));
  80.             }
  81.             oper.pop();
  82.         }
  83.         else if(str[i] == '-' && (last == 0 || last == '(')){
  84.             num.push(0.0);
  85.             oper.push('-');
  86.         }
  87.         else if(priv[str[i]] > 0){
  88.             while(oper.size() > 0 && priv[str[i]] >= priv[oper.top()]){
  89.                 y = num.top(); num.pop();
  90.                 x = num.top(); num.pop();
  91.                 char op = oper.top();
  92.                 oper.pop();
  93.                 num.push(calc(x, y, op));
  94.             }
  95.             oper.push(str[i]);
  96.         }else continue;
  97.         last = str[i];
  98.     }
  99.     while(oper.size() > 0){
  100.         y = num.top(); num.pop();
  101.         x = num.top(); num.pop();
  102.         char op = oper.top();
  103.         oper.pop();
  104.         num.push(calc(x, y, op));
  105.     }
  106.     return num.top();
  107. }
  108.  
  109. map<int, bool> mop;
  110. vector<int> num;
  111. vector<char> sym;
  112. int lft, rgt, zero, divi, typ;
  113.  
  114. const string sta[6][6] ={
  115.     {"=", "<", ">", "!=", "<=", ">="},
  116.     {"4", "+", "-", "(", "(", ")"},
  117.     {"0", "/", "/", "/", "8", "+"},
  118.     {"2", "3", "4", "5", "-", ")"},
  119.     {"+", "-", "*", "/", "1", "9"},
  120.     {"6", "7", "+", "-", "(", ")"}
  121. };
  122.  
  123. void ask(int o)
  124. {
  125.     char ch;
  126.     if(o == 2){
  127.         puts("2");
  128.         fflush(stdout);
  129.         scanf(" %c", &ch);
  130.         //ch = sta[1][rand() % 6][0];
  131.         //cout << "A: " << ch;
  132.         if(ch == '(') lft ++;
  133.         else if(ch == ')') rgt ++;
  134.         else if(ch == '4') {
  135.             num.push_back(ch);
  136.             if(mop[ch] == 0){
  137.                 mop[ch] = 1;
  138.                 typ ++;
  139.             }
  140.         }
  141.         else sym.push_back(ch);
  142.     }
  143.     else if(o == 3){
  144.         puts("3");
  145.         fflush(stdout);
  146.         scanf(" %c", &ch);
  147.         //ch = sta[2][rand() % 6][0];
  148.         //cout << "A: " << ch;
  149.  
  150.         if(ch == '0') zero ++;
  151.         else if(ch == '8') {
  152.             num.push_back(ch);
  153.             if(mop[ch] == 0){
  154.                 mop[ch] = 1;
  155.                 typ ++;
  156.             }
  157.         }
  158.         else if(ch == '/') divi ++;
  159.         else sym.push_back(ch);
  160.     }
  161.     else if(o == 4){
  162.         puts("4");
  163.         fflush(stdout);
  164.         //ch = sta[3][rand() % 6][0];
  165.         //cout << "A: " << ch;
  166.  
  167.         scanf(" %c", &ch);
  168.         if(ch == '-') sym.push_back(ch);
  169.         else if(ch == ')') rgt ++;
  170.         else {
  171.             num.push_back(ch);
  172.             if(mop[ch] == 0){
  173.                 mop[ch] = 1;
  174.                 typ ++;
  175.             }
  176.         }
  177.     }
  178. }
  179.  
  180. string ans;
  181. string s;
  182. string ss;
  183.  
  184. int sgn(double x)
  185. {
  186.     if(fabs(x) < eps) return 0;
  187.     return x > 0 ? 1 : -1;
  188. }
  189.  
  190. bool choose()
  191. {
  192.     s = "";
  193.     int DIV = divi, ZERO = zero;
  194.     int n = num.size(), m = sym.size() + DIV;
  195.     int bas = (n + ZERO) / (m + 1);
  196.     int mo = (n + ZERO) - (m + 1) * bas;
  197.     int t = 0, j = -1, k = -1;
  198.     for(int i = 0; i < m + 1; i ++){
  199.         if(mo){
  200.             if(j + 1 >= n) return 0; // 必须得放前导0了
  201.             s += num[++ j]; ++ t;
  202.             while(t <= bas && ZERO){
  203.                 s += '0'; t ++; ZERO --;
  204.             }
  205.             while(t <= bas){
  206.                 s += num[++ j]; t ++;
  207.             }
  208.             mo --;
  209.             t = 0;
  210.         }
  211.         else{
  212.             if(j + 1 >= n) return 0;
  213.             s += num[++ j]; ++ t;
  214.             while(t < bas && ZERO){
  215.                 s += '0'; t ++; ZERO --;
  216.             }
  217.             while(t < bas){
  218.                 s += num[++ j]; t ++;
  219.             }
  220.             t = 0;
  221.         }
  222.         if(DIV){
  223.             s += '/';
  224.             DIV --;
  225.         }
  226.         else{
  227.             if(k + 1 < sym.size())s += sym[++ k];
  228.         }
  229.     }
  230.     return 1;
  231. }
  232.  
  233. bool check(int o)
  234. {
  235.     if(o == 0){
  236.         zero -= 2; lft --; rgt --; divi --;
  237.         if(!choose()) {zero += 2; lft ++; rgt ++; divi ++; return 0;}
  238.         if(calculate(s) != 0){
  239.             ans = "";
  240.             for(int i = 0; i < lft; i ++) ans += '(';
  241.             ans += '0';
  242.             for(int i = 0; i < rgt; i ++) ans += ')';
  243.             ans += "=0/(" + s + ')';
  244.         }
  245.         else {zero += 2; lft ++; rgt ++; divi ++; return 0;}
  246.     }
  247.     else{
  248.         zero --;
  249.         if(!choose()) {zero ++;return 0;}
  250.         double tmp = calculate(s);
  251.         if(sgn(tmp) != 0){
  252.             if(ss[0] == '<' && sgn(tmp) > 0 || ss[0] == '>' && sgn(tmp) < 0){
  253.                 ans = "";
  254.                 for(int i = 0; i < lft; i ++) ans += '(';
  255.                 ans += '0';
  256.                 for(int i = 0; i < lft; i ++) ans += ')';
  257.                 ans += ss + s;
  258.             }
  259.             else{
  260.                 ans = "";
  261.                 for(int i = 0; i < lft; i ++) ans += '(';
  262.                 ans += s;
  263.                 for(int i = 0; i < lft; i ++) ans += ')';
  264.                 ans += ss + '0';
  265.             }
  266.         }
  267.         else if(ss == "!="){
  268.             ans = "";
  269.             for(int i = 0; i < lft; i ++) ans += '(';
  270.             ans += '0';
  271.             for(int i = 0; i < lft; i ++) ans += ')';
  272.             ans += ss + s;
  273.  
  274.         }
  275.         else {zero ++; return 0;}
  276.     }
  277.     return 1;
  278. }
  279.  
  280. int main()
  281. {
  282.     srand(time(NULL));
  283.     //fre();
  284.     puts("1");
  285.     fflush(stdout);
  286.     cin >> ss;
  287.     //ss = sta[0][rand() % 6];
  288.     //cout << ss << endl;
  289.     lft = rgt = zero = divi = typ = 0;
  290.     int tim = 0;
  291.     while(1){
  292.         if(++ tim == 1000){
  293.             int go = 1;
  294.         }
  295.         if(zero < (ss[0] == '=' ? 2 : 1) || divi == 0){
  296.             ask(3);
  297.         }
  298.         else if(lft < rgt || lft == 0){
  299.             ask(2);
  300.         }
  301.         else if(num.size() - 2 < sym.size() || rgt < lft){
  302.             ask(4);
  303.         }
  304.         else {
  305.             if(check(ss[0] == '=' ? 0 : 1)) break;
  306.             ask(4);
  307.         }
  308.     }
  309.     printf("0 ");
  310.     cout << ans << endl;
  311.     fflush(stdout);
  312.     return 0;
  313. }
  314.  
  315. /*
  316.  
  317.  题意:
  318.  
  319.  类型:
  320.  
  321.  分析:
  322.  
  323.  优化:
  324.  
  325.  trick:
  326.  
  327.  数据:
  328.  
  329.  Sample Input
  330.  
  331.  Sample Output
  332.  
  333.  >=
  334.  + / / + 8 0 ) ( ) ( - - 5 5 5 - 4 4 4 - 5 3
  335.  
  336.  */

　　

F. Flight trip

留坑。

G. Glasses with solutions

找出的子集需要满足$b\sum m-a\sum t=0$，折半搜索即可。

  1. #include<cstdio>
  2. #include<map>
  3. using namespace std;
  4. typedef long long ll;
  5. const int N=40;
  6. int n,m,A,B,i,x,y,a[N];map<ll,ll>f;
  7. ll ans;
  8. void dfsl(int x,ll y){
  9.     if(x==m){f[y]++;return;}
  10.     dfsl(x+1,y+a[x]);
  11.     dfsl(x+1,y);
  12. }
  13. void dfsr(int x,ll y){
  14.     if(x==n){ans+=f[-y];return;}
  15.     dfsr(x+1,y+a[x]);
  16.     dfsr(x+1,y);
  17. }
  18. int main(){
  19.     freopen("input.txt","r",stdin);
  20.     freopen("output.txt","w",stdout);
  21.     scanf("%d%d%d",&n,&A,&B);
  22.     for(i=0;i<n;i++){
  23.         scanf("%d%d",&x,&y);
  24.         a[i]=x*B-A*y;
  25.     }
  26.     m=n/2;
  27.     dfsl(0,0);
  28.     dfsr(m,0);
  29.     ans--;
  30.     printf("%lld",ans);
  31. }

　　

H. Hamburgers

对于每个汉堡，$O(2^6)$枚举所有它可以满足的口味，然后对于每个人判断是否满足即可。

时间复杂度$O(m\sum a+2^6\sum b)$。

  1. #include<cstdio>
  2. #include<cstring>
  3. #include<vector>
  4. using namespace std;
  5. const int N=55555;
  6. int n,m,i,j,k,x,ans[N],mx[N],b[N];
  7. vector<int>a[N];
  8. bool v[(1<<26)+5];
  9. inline int get(){
  10.     static char s[1000];
  11.     scanf("%s",s);
  12.     int t=0,len=strlen(s);
  13.     for(int i=0;i<len;i++)t|=1<<(s[i]-'a');
  14.     return t;
  15. }
  16. inline void make(int x,int y){
  17.     for(int i=x;i;i=(i-1)&x)v[i]=y;
  18. }
  19. int main(){
  20.     freopen("input.txt","r",stdin);
  21.     freopen("output.txt","w",stdout);
  22.     scanf("%d",&n);
  23.     for(i=1;i<=n;i++){
  24.         scanf("%d",&k);
  25.         while(k--){
  26.             x=get();
  27.             a[i].push_back(x);
  28.         }
  29.         ans[i]=1;
  30.     }
  31.     scanf("%d",&m);
  32.     for(i=1;i<=m;i++){
  33.         scanf("%d",&k);
  34.         for(j=1;j<=k;j++)b[j]=get();
  35.         for(j=1;j<=k;j++)make(b[j],1);
  36.         for(j=1;j<=n;j++){
  37.             int t=0;
  38.             for(x=0;x<a[j].size();x++)if(v[a[j][x]])t++;
  39.             if(t>mx[j])mx[j]=t,ans[j]=i;
  40.         }
  41.         for(j=1;j<=k;j++)make(b[j],0);
  42.     }
  43.     for(i=1;i<=n;i++)printf("%d\n",ans[i]);
  44. }
  45. /*
  46. 5
  47. 1
  48. a
  49. 3
  50. a
  51. ba
  52. cb
  53. 3
  54. vba
  55. d
  56. ba
  57. 3
  58. ca
  59. da
  60. da
  61. 3
  62. as
  63. ba
  64. af
  65. 2
  66. 4
  67. abc
  68. abcd
  69. a
  70. a
  71. 2
  72. abcdef
  73. abcdev
  74. */

　　

I. Intricate path

留坑。

J. Jumps through the Hyperspace

显然到每个点的时间越早越好，对于每种传送门预处理出每种余数下的最优等待时间即可。

  1. #include<cstdio>
  2. #include<algorithm>
  3. #include<vector>
  4. #include<queue>
  5. using namespace std;
  6. typedef pair<int,int>P;
  7. const int N=2010,inf=~0U>>1;
  8. int n,m,st,i,j,a[N],b[N],c[N],d[N],w[N][N];
  9. int f[N];
  10. char g[N][N],op[N];
  11. priority_queue<P,vector<P>,greater<P> >q;
  12. inline void ext(int x,int y){
  13.     if(f[x]>y)q.push(P(f[x]=y,x));
  14. }
  15. int main(){
  16.     freopen("input.txt","r",stdin);
  17.     freopen("output.txt","w",stdout);
  18.     scanf("%d%d%d",&n,&m,&st);
  19.     while(m--){
  20.         scanf("%s",op);
  21.         int o=op[0];
  22.         scanf("%d%d%d%d",&a[o],&b[o],&c[o],&d[o]);
  23.         for(i=0;i<c[o];i++){
  24.             w[o][i]=inf;
  25.             for(j=0;j<c[o];j++)w[o][i]=min(w[o][i],(a[o]*(i+j)+b[o])%c[o]+d[o]+j);
  26.         }
  27.     }
  28.     for(i=1;i<=n;i++)scanf("%s",g[i]+1);
  29.     for(i=1;i<=n;i++)f[i]=inf;
  30.     ext(1,st);
  31.     while(!q.empty()){
  32.         P t=q.top();q.pop();
  33.         if(f[t.second]<t.first)continue;
  34.         for(i=1;i<=n;i++)if(g[t.second][i]!='.')
  35.             ext(i,t.first+w[g[t.second][i]][t.first%c[g[t.second][i]]]);
  36.     }
  37.     if(f[n]==inf)puts("-1");else
  38.     printf("%d",f[n]-st);
  39. }

　　

K. King’s island

暴力搜索出一个上凸壳，然后翻转拼接起来得到完整的多边形即可。

  1. #include<stdio.h>
  2. #include<iostream>
  3. #include<string.h>
  4. #include<string>
  5. #include<ctype.h>
  6. #include<math.h>
  7. #include<set>
  8. #include<map>
  9. #include<vector>
  10. #include<queue>
  11. #include<bitset>
  12. #include<algorithm>
  13. #include<time.h>
  14. using namespace std;
  15. void fre() { freopen("c://test//input.in", "r", stdin); freopen("c://test//output.out", "w", stdout); }
  16. #define MS(x, y) memset(x, y, sizeof(x))
  17. #define ls o<<1
  18. #define rs o<<1|1
  19. typedef long long LL;
  20. typedef unsigned long long UL;
  21. typedef unsigned int UI;
  22. template <class T1, class T2>inline void gmax(T1 &a, T2 b) { if (b > a)a = b; }
  23. template <class T1, class T2>inline void gmin(T1 &a, T2 b) { if (b < a)a = b; }
  24. const int N = 1010, M = 0, Z = 1e9 + 7, inf = 0x3f3f3f3f;
  25. template <class T1, class T2>inline void gadd(T1 &a, T2 b) { a = (a + b) % Z; }
  26. int casenum, casei;
  27. int n;
  28. struct B
  29. {
  30.     int x, y;
  31.     bool operator < (const B & b)const
  32.     {
  33.         //y/x > b.y/b.x
  34.         return y * b.x > x * b.y;
  35.     }
  36. };
  37. B p[100000]; int g;
  38. int f[N][N][62];
  39. struct A
  40. {
  41.     int x, y, z;
  42. };
  43. bool bingo[60];
  44. vector<B>ans[60];
  45.  
  46. void getvt(int x, int y, int v, vector<B>&vt, int DX, int DY)
  47. {
  48.     if(v == 0)return;
  49.     int o = f[x][y][v];
  50.     if(f[x - p[o].x][y - p[o].y][v] == o)
  51.     {
  52.         getvt(x - p[o].x, y - p[o].y, v, vt, DX + p[o].x, DY + p[o].y);
  53.     }
  54.     else
  55.     {
  56.         vt.push_back({p[o].x + DX, p[o].y + DY});
  57.         getvt(x - p[o].x, y - p[o].y, v - 1, vt, 0, 0);
  58.     }
  59. }
  60.  
  61. void init()
  62. {
  63.     for(int i = 1; i <= 200; ++i)
  64.     {
  65.         for(int j = i + 1; j <= 200; ++j)if(__gcd(i, j) == 1)
  66.         {
  67.             int W = (i * i + j * j);
  68.             int w = sqrt(W);
  69.             if(w * w == W)
  70.             {
  71.                 p[++g] = {i, j};
  72.                 p[++g] = {j, i};
  73.             }
  74.         }
  75.     }
  76.     sort(p + 1, p + g + 1);
  77.     //printf("%d\n", g);
  78.  
  79.     MS(f, 0);
  80.     f[0][0][0] = 1;
  81.     for(int i = 0; i <= 500; ++i)
  82.     {
  83.         for(int j = 0; j <= 500; ++j)
  84.         {
  85.             for(int v = 29; v >= 0; --v)if(f[i][j][v])
  86.             {
  87.                 //
  88.                 int k = f[i][j][v];
  89.                 auto &it = f[i + p[k].x][j + p[k].y][v];
  90.                 if(it == 0)it = k;
  91.                 //
  92.                 for(int k = f[i][j][v] + 1; k <= g; ++k)
  93.                 {
  94.                     auto &it = f[i + p[k].x][j + p[k].y][v + 1];
  95.                     if(it == 0)it = k;
  96.                 }
  97.                 if(i == 0 || j == 0)continue;
  98.                 for(int u = 1; u + v <= 30; ++u)if(f[i][j][u] && f[i][j][u] != f[i][j][v])
  99.                 {
  100.                     if(!bingo[u + v])
  101.                     {
  102.                         bingo[u + v] = 1;
  103.                         vector<B>up, down;
  104.                         getvt(i, j, u, up, 0, 0);
  105.                         getvt(i, j, v, down, 0, 0);
  106.                         int x = i;
  107.                         int y = j;
  108.                         //printf("%d %d %d\n", u + v, up.size(), down.size());
  109.                         //puts("up------------");
  110.                         for(int r = 0; r < up.size(); ++r)
  111.                         {
  112.                             x -= up[r].x;
  113.                             y -= up[r].y;
  114.                             ans[u + v].push_back({x, y});
  115.                             //printf("%d %d\n", up[r].x, up[r].y);
  116.                         }
  117.                         //puts("down------------");
  118.                         for(int r = 0; r < down.size(); ++r)
  119.                         {
  120.                             x += down[r].x;
  121.                             y += down[r].y;
  122.                             ans[u + v].push_back({x, y});
  123.                             //printf("%d %d\n", down[r].x, down[r].y);
  124.                         }
  125.                         //puts("ans------");
  126.                         for(auto w : ans[u + v])
  127.                         {
  128.                             //printf("%d %d\n", w.x, w.y);
  129.                         }
  130.                         int pasue = 1;
  131.                     }
  132.                 }
  133.             }
  134.         }
  135.     }
  136.     for(int i = 1;i <= 30; ++i)
  137.     {
  138.         //printf("%d\n", bingo[i]);
  139.     }
  140. }
  141. int main()
  142. {
  143.     freopen("input.txt", "r", stdin);
  144.     freopen("output.txt", "w", stdout);
  145.     init();
  146. 	while(~scanf("%d", &n))
  147. 	{
  148. 	    if(n == 3)
  149.         {
  150.             puts("0 0\n4 3\n-20 21");
  151.             continue;
  152.         }
  153.         for(auto it : ans[n])
  154.         {
  155.             printf("%d %d\n", it.x, it.y);
  156.         }
  157. 	}
  158.  
  159. 	return 0;
  160. }
  161. /*
  162. 【trick&&吐槽】
  163.  
  164. 【题意】
  165.  
  166. 【分析】
  167.  
  168. 【时间复杂度&&优化】
  169. 28 195
  170. 9 15
  171. -116 178
  172. 4
  173. 20 21
  174. 1 -159
  175. -236 -11
  176. -92 6
  177. 5
  178. 20 50
  179. 3 -94
  180. -16 -274
  181. -236 18
  182. -92 35
  183.  
  184. */

　　

L. Lexica

无视同一行/同一列的两个格子的位置关系，只需要在它们不相同时将方案数乘$2$，如此一来只关心每一行/每一列有多少格子没满足。

设$f[i][j][S][k]$表示考虑到$(i,j)$，每一列剩余情况为$S$，第$i$行还有$k$个格子需要填充时的方案数，然后逐格转移即可。

时间复杂度$O(n3^n)$。

  1. #include<cstdio>
  2. typedef unsigned long long ll;
  3. const int N=18,M=1600000;
  4. int n,m,all,i,j,k,x,y,z,A,B,C,S,o,p[N],mask,cnt[N];
  5. char a[N][N];
  6. int e;
  7. ll f[2][M][3],mul;
  8. int w[M];
  9. inline void clr(){
  10.     for(int i=0;i<=mask;i++)for(int j=0;j<3;j++)f[e^1][i][j]=0;
  11. }
  12. inline void nxt(){
  13.     //for(int i=0;i<=mask;i++)for(int j=0;j<3;j++)f[i][j]=g[i][j];
  14. }
  15. inline int get(int S,int x){
  16.     return S/p[x]%3;
  17. }
  18. void dfs(int x,int y,int z){
  19. 		if(x==n){
  20. 			w[y]=z;
  21. 			return;
  22. 		}
  23. 		for(int i=0;i<3;i++)dfs(x+1,y*3+i,z*2+(!!i));
  24. }
  25. int main(){
  26.     freopen("input.txt","r",stdin);
  27.     freopen("output.txt","w",stdout);
  28.     scanf("%d",&n);
  29.     m=n+2;
  30.     for(i=0;i<m;i++)scanf("%s",a[i]);
  31.     mul=1;
  32.     for(p[0]=i=1;i<N;i++)p[i]=p[i-1]*3;
  33.     all=p[n];
  34.     for(i=1;i<=n;i++){
  35.         if(a[0][i]!='#'&&a[m-1][i]!='#'&&a[0][i]!=a[m-1][i])mul*=2;
  36.         if(a[i][0]!='#'&&a[i][m-1]!='#'&&a[i][0]!=a[i][m-1])mul*=2;
  37.         if(a[i][0]!='#')cnt[i]++;
  38.         if(a[i][m-1]!='#')cnt[i]++;
  39.         if(a[0][i]!='#')mask+=p[i-1];
  40.         if(a[m-1][i]!='#')mask+=p[i-1];
  41.     }
  42.     dfs(0,0,0);
  43.     f[0][mask][0]=mul;
  44.     for(i=1;i<=n;i++){
  45.         clr();
  46.         int o=cnt[i];
  47.         for(A=0;A<=mask;A++)f[e^1][A][o]=f[e][A][0];
  48.         e^=1;
  49.         for(j=0;j<n;j++)if(a[i][j+1]=='.'){
  50.             clr();
  51.             for(S=0;S<=mask;S++){
  52.             		if(w[S]>>j&1)for(k=0;k<=o;k++)if(f[e][S][k])f[e^1][S-p[j]][k]+=f[e][S][k];
  53.             		for(k=1;k<=o;k++)if(f[e][S][k])f[e^1][S][k-1]+=f[e][S][k];
  54.             }
  55.             e^=1;
  56.         }
  57.     }
  58.     printf("%llu",f[e][0][0]);
  59. }
  60. /*
  61. 5
  62. ##ARRR#
  63. H.#.##S
  64. Y.....E
  65. O.#.#.E
  66. ##....E
  67. ###.#.E
  68. #XNT#A#
  69. */

　　

M. Maximal paths

树形DP求出每个点内部从叶子往上、从上往下的最大数字串即可。需要手写高精度。

  1. #include<cstdio>
  2. #include<algorithm>
  3. using namespace std;
  4. typedef unsigned long long ll;
  5. typedef __int128 lll;
  6. const int N=10000010,M=50;
  7. const ll MO=1000000000000000000ULL;
  8. int n,i;char a[N];
  9. struct P{
  10.     ll v[3];
  11.     P(){v[0]=v[1]=v[2]=0;}
  12.     P operator+(int b){
  13.         P c;
  14.         for(int i=0;i<3;i++)c.v[i]=v[i];
  15.         c.v[0]+=b;
  16.         for(int i=0;i<2;i++)if(c.v[i]>=MO){
  17.             c.v[i+1]+=c.v[i]/MO;
  18.             c.v[i]-=MO;
  19.         }
  20.         return c;
  21.     }
  22.     P operator+(const P&b){
  23.         P c;
  24.         for(int i=0;i<3;i++)c.v[i]=v[i]+b.v[i];
  25.         for(int i=0;i<2;i++)if(c.v[i]>=MO){
  26.             c.v[i+1]+=c.v[i]/MO;
  27.             c.v[i]-=MO;
  28.         }
  29.         return c;
  30.     }
  31.     P operator*(int b){
  32.         P c;
  33.         for(int i=0;i<3;i++)c.v[i]=v[i]*b;
  34.         for(int i=0;i<2;i++)c.v[i+1]+=c.v[i]/MO,c.v[i]%=MO;
  35.         c.v[2]%=MO;
  36.         return c;
  37.     }
  38.     P operator*(const P&b){
  39.         P c;
  40.         static lll f[3];
  41.         f[0]=f[1]=f[2]=0;
  42.         for(int i=0;i<3;i++)if(v[i])for(int j=0;i+j<3;j++)if(b.v[j])f[i+j]+=(lll)v[i]*b.v[j];
  43.         for(int i=0;i<2;i++)if(f[i]>=MO)f[i+1]+=f[i]/MO,f[i]%=MO;
  44.         if(f[2]>=MO)f[2]%=MO;
  45.         for(int i=0;i<3;i++)c.v[i]=f[i];
  46.         return c;
  47.     }
  48.     void up(const P&b){//max=
  49.         int i;
  50.         for(i=2;~i;i--)if(b.v[i]!=v[i])break;
  51.         if(i<0)return;
  52.         if(b.v[i]<v[i])return;
  53.         for(i=0;i<3;i++)v[i]=b.v[i];
  54.     }
  55.     void write(){
  56.         int i=2;
  57.         while(i&&!v[i])i--;
  58.         printf("%llu",v[i]);
  59.         for(int j=i-1;~j;j--)printf("%018llu",v[j]);
  60.         puts("");
  61.     }
  62. }f[M],g[M],p[M],ans,ff,gg,mx[M];
  63. int d[M];
  64. unsigned int seed,base;
  65. void dfs(int x,int k){
  66.     f[k]=g[k]=mx[k]=P();
  67.     d[k]=0;
  68.     for(int i=0;i<2;i++){
  69.         int y=x<<1|i;
  70.         if(y>n)continue;
  71.         dfs(y,k+1);
  72.         ff=(f[k+1]*10)+((int)a[y]);
  73.         gg=g[k+1]+(p[d[k+1]]*((int)a[y]));
  74.         mx[k].up((ff*p[d[k]])+g[k]);
  75.         mx[k].up((f[k]*p[d[k+1]+1])+gg);
  76.         /*if(x==1){
  77.             puts("debug");
  78.             ff.write();
  79.             f[k].write();
  80.         }*/
  81.         f[k].up(ff);
  82.         g[k].up(gg);
  83.         d[k]=max(d[k],d[k+1]+1);
  84.         /*if(x==1){
  85.             puts("debug");
  86.             ff.write();
  87.             f[k].write();
  88.         }*/
  89.     }
  90.    /* printf("%d:\n",x);
  91.     f[k].write();
  92.     g[k].write();
  93.     mx[k].write();*/
  94.     ans=ans+mx[k];
  95. }
  96. int main(){
  97.     freopen("input.txt","r",stdin);
  98.     freopen("output.txt","w",stdout);
  99.     scanf("%d%u",&n,&seed);
  100.     base=(1U<<31)-1;
  101.     p[0].v[0]=1;
  102.     for(i=1;i<M;i++)p[i]=p[i-1]*10;
  103.     for(i=1;i<=n;i++){
  104.         a[i]=((seed&base)>>16)%9+1;
  105.         seed=seed*1103515245+12345;
  106.         //printf("%d\n",a[i]); i and i/2
  107.     }
  108.     dfs(1,0);
  109.     ans.write();
  110. }

　　

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