2018 AICCSA Programming Contest

A Tree Game

B Rectangles

思路:如果存在大于0的交面积的话, 那么肯定能找到一条水平的直线 和 一条垂直的直线,

使得水平直线的左右两边点的个数相等且为n, 垂直直线的左右两边点的个数相等且为n

也就是说不能有点在这两条线上, 否则交面积为0

然后左上角的点和右下角的点配对, 左下角的点和右上角的点配对

代码:

  1. #pragma GCC optimize(2)
  2. #pragma GCC optimize(3)
  3. #pragma GCC optimize(4)
  4. #include<bits/stdc++.h>
  5. using namespace std;
  6. #define fi first
  7. #define se second
  8. #define pi acos(-1.0)
  9. #define LL long long
  10. //#define mp make_pair
  11. #define pb push_back
  12. #define ls rt<<1, l, m
  13. #define rs rt<<1|1, m+1, r
  14. #define ULL unsigned LL
  15. #define pll pair<LL, LL>
  16. #define pli pair<LL, int>
  17. #define pii pair<int, int>
  18. #define piii pair<pii, int>
  19. #define mem(a, b) memset(a, b, sizeof(a))
  20. #define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
  21. #define fopen freopen("in.txt", "r", stdin);freopen("out.txt", "w", stout);
  22. //head
  23.  
  24. const int N = 2e5 + , M = 1e5 + ;
  25. const int MOD = 1e9 + ;
  26. pii a[N];
  27. int fac[M];
  28. bool cmp(pii a, pii b) {
  29.     return a.se < b.se;
  30. }
  31. void init() {
  32.     fac[] = ;
  33.     for (int i = ; i < M; i++) fac[i] = (1LL * fac[i-] * i) % MOD;
  34. }
  35. int main() {
  36.     int T, n;
  37.     init();
  38.     scanf("%d", &T);
  39.     while(T--) {
  40.         scanf("%d", &n);
  41.         for (int i = ; i <= *n; i++) scanf("%d %d", &a[i].fi, &a[i].se);
  42.         bool f = false;
  43.         sort(a+, a++*n);
  44.         double x = , y = ;
  45.         if(a[n].fi != a[n+].fi) x = (a[n].fi + a[n+].fi) / 2.0;
  46.         else f = true;
  47.  
  48.         sort(a+, a++*n, cmp);
  49.         if(a[n].se != a[n+].se) y = (a[n].se + a[n+].se) / 2.0;
  50.         else f = true;
  51.         int cnt = ;
  52.         for (int i = ; i <= *n; i++) if(a[i].fi > x && a[i].se > y) cnt++;
  53.         if(f) printf("0\n");
  54.         else printf("%lld\n", (1LL * fac[cnt] * fac[n-cnt]) % MOD);
  55.     }
  56.     return ;
  57. }

C Function

思路:打表找规律, 发现ai的系数为C(n+1, i) - 1

代码:

  1. #pragma GCC optimize(2)
  2. #pragma GCC optimize(3)
  3. #pragma GCC optimize(4)
  4. #include<bits/stdc++.h>
  5. using namespace std;
  6. #define fi first
  7. #define se second
  8. #define pi acos(-1.0)
  9. #define LL long long
  10. //#define mp make_pair
  11. #define pb push_back
  12. #define ls rt<<1, l, m
  13. #define rs rt<<1|1, m+1, r
  14. #define ULL unsigned LL
  15. #define pll pair<LL, LL>
  16. #define pli pair<LL, int>
  17. #define pii pair<int, int>
  18. #define piii pair<pii, int>
  19. #define mem(a, b) memset(a, b, sizeof(a))
  20. #define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
  21. #define fopen freopen("in.txt", "r", stdin);freopen("out.txt", "w", stout);
  22. //head
  23.  
  24. const int N = 1e6 + ;
  25. const int MOD = 1e9 + ;
  26. int a[N];
  27. LL fac[N], inv[N];
  28. LL q_pow(LL n, LL k) {
  29.     LL ans = ;
  30.     while(k) {
  31.         if(k&) ans = (ans * n) % MOD;
  32.         n = (n * n) % MOD;
  33.         k >>= ;
  34.     }
  35.     return ans;
  36. }
  37. void init() {
  38.     fac[] = ;
  39.     for (int i = ; i < N; i++) fac[i] = fac[i-] * i % MOD;
  40.     inv[N-] = q_pow(fac[N-], MOD-);
  41.     for (int i = N-; i >= ; i--) inv[i] = inv[i+] * (i+) % MOD;
  42. }
  43. LL C(int n, int m) {
  44.     return fac[n] * inv[m] % MOD * inv[n-m] % MOD;
  45. }
  46. int main() {
  47.     int T;
  48.     init();
  49.     scanf("%d", &T);
  50.     while(T--) {
  51.         int n;
  52.         scanf("%d", &n);
  53.         for (int i = ; i <= n; i++) scanf("%d", &a[i]);
  54.         LL ans = ;
  55.         for (int i = ; i <= n; i++) {
  56.             (ans = ans + (C(n+, i) - ) * a[i] % MOD) %= MOD;
  57.         }
  58.         printf("%lld\n", (ans + MOD) % MOD);
  59.     }
  60.     return ;
  61. }

D Two Sequences

思路:将a数组放到集合里, 方便查找删除

代码:

  1. #pragma GCC optimize(2)
  2. #pragma GCC optimize(3)
  3. #pragma GCC optimize(4)
  4. #include<bits/stdc++.h>
  5. using namespace std;
  6. #define fi first
  7. #define se second
  8. #define pi acos(-1.0)
  9. #define LL long long
  10. //#define mp make_pair
  11. #define pb push_back
  12. #define ls rt<<1, l, m
  13. #define rs rt<<1|1, m+1, r
  14. #define ULL unsigned LL
  15. #define pll pair<LL, LL>
  16. #define pli pair<LL, int>
  17. #define pii pair<int, int>
  18. #define piii pair<pii, int>
  19. #define mem(a, b) memset(a, b, sizeof(a))
  20. #define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
  21. #define fopen freopen("in.txt", "r", stdin);freopen("out.txt", "w", stout);
  22. //head
  23.  
  24. const int N = 1e5 + ;
  25. int a[N], b[N];
  26. multiset<int> s;
  27. vector<int> vc;
  28. int main() {
  29.     int T, n, k;
  30.     scanf("%d", &T);
  31.     while(T--) {
  32.         scanf("%d %d", &n, &k);
  33.         s.clear();
  34.         vc.clear();
  35.         for (int i = ; i <= n; i++) scanf("%d", &a[i]), s.insert(a[i]);
  36.         for (int i = ; i <= n; i++) scanf("%d", &b[i]);
  37.         for (int i = ; i <= n; i++) {
  38.             multiset<int>:: iterator it = s.lower_bound(b[i]);
  39.             if(it == s.end() || *it != b[i]) {
  40.                 vc.pb(b[i]);
  41.             }
  42.             else s.erase(it);
  43.         }
  44.         if((int)vc.size() == ) puts("YES");
  45.         else if((int)vc.size() == ) {
  46.             if(*s.begin() - k <= vc[] && vc[] <= *s.begin() + k) puts("YES");
  47.             else puts("NO");
  48.         }
  49.         else puts("NO");
  50.     }
  51.     return ;
  52. }

E Connecting Components

F Mirror

G TeddyBearsDay

思路:对于每个点, 它连向父亲的边只有当它的子树中不能自销的多余部分才会用到

代码:

  1. #pragma GCC optimize(2)
  2. #pragma GCC optimize(3)
  3. #pragma GCC optimize(4)
  4. #include<bits/stdc++.h>
  5. using namespace std;
  6. #define fi first
  7. #define se second
  8. #define pi acos(-1.0)
  9. #define LL long long
  10. //#define mp make_pair
  11. #define pb push_back
  12. #define ls rt<<1, l, m
  13. #define rs rt<<1|1, m+1, r
  14. #define ULL unsigned LL
  15. #define pll pair<LL, LL>
  16. #define pli pair<LL, int>
  17. #define pii pair<int, int>
  18. #define piii pair<pii, int>
  19. #define mem(a, b) memset(a, b, sizeof(a))
  20. #define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
  21. #define fopen freopen("in.txt", "r", stdin);freopen("out.txt", "w", stout);
  22. //head
  23.  
  24. const int N = 1e4 + ;
  25. vector<pii> g[N];
  26. int send[N], rev[N];
  27. LL ans = ;
  28.  
  29. pii dfs(int u, int o, int w) {
  30.     pii tmp = {send[u], rev[u]};
  31.     for (pii p : g[u]) {
  32.         if(p.fi != o) {
  33.              pii pp = dfs(p.fi, u, p.se);
  34.              tmp.fi += pp.fi;
  35.              tmp.se += pp.se;
  36.         }
  37.     }
  38.     ans += 1LL * abs(tmp.fi - tmp.se) * w;
  39.     return tmp;
  40. }
  41. int main() {
  42.     int T, n, u, v, w, q;
  43.     scanf("%d", &T);
  44.     while(T--) {
  45.         scanf("%d", &n);
  46.         for (int i = ; i <= n; i++) g[i].clear(), send[i] = rev[i] = ;
  47.         for (int i = ; i < n; i++) {
  48.             scanf("%d %d %d", &u, &v, &w);
  49.             g[u].pb({v, w});
  50.             g[v].pb({u, w});
  51.         }
  52.         scanf("%d", &q);
  53.         while(q--) {
  54.             scanf("%d %d", &u, &v);
  55.             send[u]++;
  56.             rev[v]++;
  57.         }
  58.         ans = ;
  59.         dfs(, , );
  60.         printf("%lld\n", ans);
  61.     }
  62.     return ;
  63. }

H Win Strategy

思路:背包dp变形

代码:

  1. #pragma GCC optimize(2)
  2. #pragma GCC optimize(3)
  3. #pragma GCC optimize(4)
  4. #include<bits/stdc++.h>
  5. using namespace std;
  6. #define fi first
  7. #define se second
  8. #define pi acos(-1.0)
  9. #define LL long long
  10. //#define mp make_pair
  11. #define pb push_back
  12. #define ls rt<<1, l, m
  13. #define rs rt<<1|1, m+1, r
  14. #define ULL unsigned LL
  15. #define pll pair<LL, LL>
  16. #define pli pair<LL, int>
  17. #define pii pair<int, int>
  18. #define piii pair<pii, int>
  19. #define mem(a, b) memset(a, b, sizeof(a))
  20. #define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
  21. #define fopen freopen("in.txt", "r", stdin);freopen("out.txt", "w", stout);
  22. //head
  23.  
  24. const int N = 1e3 + ;
  25. pii a[N];
  26. int dp[N];
  27. int main() {
  28.     int T, n, L;
  29.     scanf("%d", &T);
  30.     while(T--) {
  31.         scanf("%d %d", &n, &L);
  32.         for (int i = ; i <= n; i++) scanf("%d %d", &a[i].fi, &a[i].se);
  33.         for (int i = ; i <= L; i++) dp[i] = ;
  34.         for (int i = ; i <= n; i++) {
  35.             for (int j = L; j >= a[i].se; j--) {
  36.                 if(j-a[i].se +  >= a[i].fi) dp[j] = max(dp[j], dp[j-a[i].se] + );
  37.             }
  38.         }
  39.         printf("%d\n", dp[L]);
  40.     }
  41.     return ;
  42. }

I Tours

思路:二分答案, check时对于每辆bus, 如果它上一天是空闲的, 才能填补今天的空缺, 然后今天原本的车就是昨天需要的车, 不够的拿昨天剩余的补

代码:

  1. #pragma GCC optimize(2)
  2. #pragma GCC optimize(3)
  3. #pragma GCC optimize(4)
  4. #include<bits/stdc++.h>
  5. using namespace std;
  6. #define fi first
  7. #define se second
  8. #define pi acos(-1.0)
  9. #define LL long long
  10. //#define mp make_pair
  11. #define pb push_back
  12. #define ls rt<<1, l, m
  13. #define rs rt<<1|1, m+1, r
  14. #define ULL unsigned LL
  15. #define pll pair<LL, LL>
  16. #define pli pair<LL, int>
  17. #define pii pair<int, int>
  18. #define piii pair<pii, int>
  19. #define mem(a, b) memset(a, b, sizeof(a))
  20. #define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
  21. #define fopen freopen("in.txt", "r", stdin);freopen("out.txt", "w", stout);
  22. //head
  23.  
  24. const int N = 1e5 + ;
  25. vector<int> a[N];
  26. int T, m, n, r, t, tmp[N];
  27. bool check(int x) {
  28.     int pre = ;
  29.     for (int i = ; i <= n; i++) {
  30.         if(i == ) {
  31.             int tot = ;
  32.             for (int j = ; j <= m; j++) {
  33.                 tmp[j] = (a[j][i] + r - ) / r;
  34.                 tot += tmp[j];
  35.             }
  36.             if(tot > x) return false;
  37.             pre = x - tot;
  38.         }
  39.         else {
  40.             int tot = ;
  41.             for (int j = ; j <= m; j++) {
  42.                 if(tmp[j] >= (a[j][i] + r - ) / r) tmp[j] = (a[j][i] + r - ) / r;
  43.                 else {
  44.                     if(pre >= (a[j][i] + r - ) / r - tmp[j]) pre -= (a[j][i] + r - ) / r - tmp[j];
  45.                     else return false;
  46.                     tmp[j] = (a[j][i] + r - ) / r;
  47.                 }
  48.                 tot += (a[j][i] + r - ) / r;
  49.             }
  50.             if(tot > x) return false;
  51.             pre = x - tot;
  52.         }
  53.     }
  54.     return true;
  55. }
  56. int main() {
  57.     scanf("%d", &T);
  58.     while(T--) {
  59.         scanf("%d %d %d", &m, &n, &r);
  60.         for (int i = ; i <= m; i++) {
  61.             a[i].clear();
  62.             a[i].pb();
  63.             for (int j = ; j <= n; j++) {
  64.                 scanf("%d", &t);
  65.                 a[i].pb(t);
  66.             }
  67.         }
  68.         int l = , r = 5e5, mid = l+r >> ;
  69.         while(l < r) {
  70.             if(check(mid)) r = mid;
  71.             else l = mid+;
  72.             mid = l+r >> ;
  73.         }
  74.         printf("%d\n", mid);
  75.     }
  76.     return ;
  77. }

J Restricted Vertex Cover

思路:2-sat

建边:

对于一条mark的边,

如果其中一点在点覆盖中, 那么另外一点肯定不在点覆盖中

如果其中一点不在点覆盖中, 那么另外一点肯定在点覆盖中

对于一条unmark的边,

如果其中一点在点覆盖中, 那么另外一点不确定

如果其中一点不在点覆盖中, 那么另外一点肯定在点覆盖中

代码:

  1. #include<bits/stdc++.h>
  2. using namespace std;
  3. const int N=;
  4. int ins[N],dfn[N],low[N],cnt,sta[N];
  5. int top,v,u;
  6. int Next[N],head[N],to[N];
  7. int tot;
  8. int scc[N];
  9. int scccnt;
  10. void make_list(int u,int v){
  11.     Next[++tot]=head[u],head[u]=tot,to[tot]=v;
  12. }
  13. void tarjan(int x){
  14.     ins[x]=dfn[x]=low[x]=++cnt,sta[top++]=x;
  15.     for(int p=head[x],v=to[p];p;p=Next[p],v=to[p])
  16.         if(!dfn[v])tarjan(v),low[x]=min(low[x],low[v]);
  17.         else if(ins[v])low[x]=min(low[x],dfn[v]);
  18.     if(low[x]==dfn[x]){
  19.         scc[x]=++scccnt,ins[x]=;
  20.         while((u=sta[--top])!=x)ins[u]=,scc[u]=scccnt;
  21.     }
  22. }
  23. int main(){
  24.     int T;
  25.     int n,m;
  26.     int u,v,w;
  27.     scanf("%d",&T);
  28.     for(int t=;t<=T;t++){
  29.         scanf("%d%d",&n,&m);
  30.         memset(head,,sizeof(int)*(n*+));
  31.         memset(dfn,,sizeof(int)*(n*+));
  32.         top=;
  33.         tot=;
  34.         memset(scc,,sizeof(int)*(n*+));
  35.         memset(low,,sizeof(int)*(n*+));
  36.         scccnt=;
  37.         memset(ins,,sizeof(int)*(n*+));
  38.         cnt=;
  39.         memset(sta,,sizeof(int)*(n*+));
  40.         for(int i=;i<=m;i++){
  41.             scanf("%d%d%d",&u,&v,&w);
  42.             if(w){
  43.                 make_list(u,v+n);
  44.                 make_list(v,u+n);
  45.                 make_list(u+n,v);
  46.                 make_list(v+n,u);
  47.             }
  48.             else{
  49.                 make_list(u+n,v);
  50.                 make_list(v+n,u);
  51.             }
  52.         }
  53.         for(int i=;i<=*n;i++){
  54.             if(!dfn[i])tarjan(i);
  55.         }
  56.         bool ok=;
  57.         for(int i=;i<=n;i++){
  58.             ok&=(scc[i]!=scc[i+n]);
  59.         }
  60.         if(ok)puts("YES");
  61.         else puts("NO");
  62.     }
  63.     return ;
  64. }

2018 AICCSA Programming Contest的更多相关文章

  1. 2018 JUST Programming Contest 1.0 题解

    题目链接  gym101778 Problem A 转化成绝对值之后算一下概率.这个题有点像 2018 ZOJ Monthly March Problem D ? 不过那个题要难一些~ #includ ...

  2. AtCoder SoundHound Inc. Programming Contest 2018 E + Graph (soundhound2018_summer_qual_e)

    原文链接https://www.cnblogs.com/zhouzhendong/p/AtCoder-SoundHound-Inc-Programming-Contest-2018-E.html 题目 ...

  3. 2018 ACM-ICPC, Syrian Collegiate Programming Contest

    2018 ACM-ICPC, Syrian Collegiate Programming Contest A Hello SCPC 2018! 水题 B Binary Hamming 水题 C Por ...

  4. 2018 German Collegiate Programming Contest (GCPC 18)

    2018 German Collegiate Programming Contest (GCPC 18) Attack on Alpha-Zet 建树,求lca 代码: #include <al ...

  5. The 2018 ACM-ICPC China JiangSu Provincial Programming Contest快速幂取模及求逆元

    题目来源 The 2018 ACM-ICPC China JiangSu Provincial Programming Contest 35.4% 1000ms 65536K Persona5 Per ...

  6. (寒假开黑gym)2018 ACM-ICPC, Syrian Collegiate Programming Contest（爽题）

    layout: post title: (寒假开黑gym)2018 ACM-ICPC, Syrian Collegiate Programming Contest(爽题) author: " ...

  7. ACM International Collegiate Programming Contest, Tishreen Collegiate Programming Contest (2018) Syria, Lattakia, Tishreen University, April, 30, 2018

    ACM International Collegiate Programming Contest, Tishreen Collegiate Programming Contest (2018) Syr ...

  8. (寒假开黑gym)2018 ACM-ICPC, Syrian Collegiate Programming Contest

    layout: post title: (寒假开黑gym)2018 ACM-ICPC, Syrian Collegiate Programming Contest author: "luow ...

  9. (寒假GYM开黑)2018-2019 ACM-ICPC Nordic Collegiate Programming Contest (NCPC 2018)

    layout: post title: 2018-2019 ACM-ICPC Nordic Collegiate Programming Contest (NCPC 2018) author: &qu ...

随机推荐

  1. mysql引擎和事务

    对于应用程序和用户来说,同样一张表的数据无论用什么引擎来存储,看到的数据都是一样的,只是不同的引擎在功能.占用空间大小.读取性能等方面可能有所差别. mysql最常用的存储引擎为Innodb.MyIS ...

  2. Linux一些基本配置

    Linux发行版:centos 6.5 配置yum源 wget http://mirrors.163.com/.help/CentOS6-Base-163.repo -P /etc/yum.repos ...

  3. 2018-2019-2 《网络对抗技术》Exp1 PC平台逆向破解 Week3 20165211

    目录 实验目标 实验基础知识准备 Linux基本操作理解 汇编指令的机器码 BOF原理 反汇编和十六进制编程器 实验内容 任务一:手工修改可执行文件 任务二:利用foo函数的Bof漏洞,触发getSh ...

  4. 福州大学第十五届程序设计竞赛_重现赛B题迷宫寻宝

    Problem B 迷宫寻宝 Accept: 52    Submit: 183Time Limit: 1000 mSec    Memory Limit : 32768 KB  Problem De ...

  5. webpack对于引入的模块无法智能代码提示

    前端模块太多了,模块里的方法比较难记住,所以我们一般靠的都是IDE的代码提示. 但是有时候我们会发现对于引入的模块没有代码提示,我也安装了模块呀,为什么没有代码提示? 主要是package.json的 ...

  6. Codeforces 711D Directed Roads - 组合数学

    ZS the Coder and Chris the Baboon has explored Udayland for quite some time. They realize that it co ...

  7. Hunter’s Apprentice 【判断多边形边界曲线顺逆时针】

    问题 H: Hunter's Apprentice 时间限制: 1 Sec  内存限制: 128 MB 提交: 353  解决: 39 [提交] [状态] [命题人:admin] 题目描述 When ...

  8. bzoj1566: [NOI2009]管道取珠 DP

    题目链接 https://www.lydsy.com/JudgeOnline/problem.php?id=1566 思路 n个球,第i个球颜色为ai,对于颜色j,对答案的贡献为颜色为j的球的个数的平 ...

  9. MATLAB小波包的分解与重构

    该文章用来直观上先感受一下小波包的分解与重构   例1 有一个信号,变量名为wave,随便找一个信号load进来就行了. t=wpdec(wave,3,'dmey'); t2 = wpjoin(t,[ ...

  10. SpringBoot 整合携程Apollo配置管理中心

    携程官网对apollo的使用讲解了很多种方式的使用,但是感觉一些细节还是没讲全,特别是eureka配置中心地址的配置 这里对springboot整合apollo说一下 >SpringBoot启动 ...