codeforce 379(div.2)

A、B略

C题 ——贪心，二分查找：

对于每一个a[i]， 在d中二分查找 s-b[i]，注意不要忘记计算速度为x时需要花费的最小时间，以及整数范围为64位整数

 1 #include <cstdio>
 2 #include <algorithm>
 3 #include <cstring>
 4
 5 using namespace std;
 6 const int maxn = 2*100000+10;
 7 typedef long long LL;
 8
 9 int n, m, k;
10 int x, s;
11 int a[maxn], b[maxn], c[maxn], d[maxn];
12
13 int main() {
14     scanf("%d%d%d", &n, &m, &k);
15     scanf("%d%d", &x, &s);
16     a[0] = x;
17     b[0] = 0;
18     for(int i = 1; i <= m; i++) scanf("%d",&a[i]);
19     for(int i = 1; i <= m; i++) scanf("%d",&b[i]);
20     for(int i = 0; i < k; i++) scanf("%d",&c[i]);
21     for(int i = 0; i < k; i++) scanf("%d",&d[i]);
22     LL  ans = LL(n)*x;
23     for(int i = 0; i <=  m; i++) {
24         if(b[i] <= s) {
25             int temp = b[i];
26             int tt = upper_bound(d, d+k, s - temp) - d;
27             if(tt) {
28                 tt--;
29                 LL  as = n - c[tt] > 0 ? LL(n - c[tt]) * a[i] : 0;
30                 ans = min(ans, as);
31             }
32             else {
33                 LL  as = LL(n) * a[i];
34                 ans = min(ans, as);
35             }
36         }
37     }
38     printf("%I64d\n", ans);
39 }

D 思路很简单，注意细节即可

  1 #include <cstdio>
  2 #include <vector>
  3 #include <algorithm>
  4 #include <cstring>
  5 #include <cstdlib>
  6 #include <cmath>
  7 using namespace std;
  8 const int maxn = 500000+10;
  9
 10 struct P{
 11     char ch;
 12     double dist;
 13     P(char c, double d): ch(c), dist(d){}
 14     bool operator < (const P& b)const {
 15         return dist < b.dist;
 16     }
 17 };
 18 vector<P> dig1, dig2, dig3, dig4;
 19 vector<P> ver1, ver2, ver3, ver4;
 20
 21 int n;
 22 int x0, y0_;
 23 #define y0  y0_
 24 void ins(char c, int dx, int dy) {
 25     if(dx == dy) {
 26         if(dx > 0) {
 27             double dis = sqrt(double(dy)*dy + double(dx)*dx);
 28             dig1.push_back(P(c, dis));
 29         }
 30         if(dx < 0) {
 31             double dis = sqrt(double(dy)*dy + double(dx)*dx);
 32             dig3.push_back(P(c, dis));
 33
 34         }
 35     }
 36     if(dx == -dy) {
 37         if(dx > 0) {
 38             double dis = sqrt(double(dy)*dy + double(dx)*dx);
 39             dig4.push_back(P(c, dis));
 40         }
 41         if(dx < 0) {
 42             double dis = sqrt(double(dy)*dy + double(dx)*dx);
 43             dig2.push_back(P(c, dis));
 44
 45         }
 46     }
 47     if(dy == 0){
 48         if(dx > 0) {
 49             double dis = sqrt(double(dy)*dy + double(dx)*dx);
 50             ver1.push_back(P(c,dis));
 51         }
 52         if(dx < 0) {
 53             double dis = sqrt(double(dy)*dy + double(dx)*dx);
 54             ver3.push_back(P(c,dis));
 55
 56         }
 57     }
 58     if(dx == 0){
 59         if(dy > 0) {
 60             double dis = sqrt(double(dy)*dy + double(dx)*dx);
 61             ver2.push_back(P(c, dis));
 62         }
 63         if(dy < 0) {
 64             double dis = sqrt(double(dy)*dy + double(dx)*dx);
 65             ver4.push_back(P (c, dis));
 66         }
 67     }
 68 }
 69 int main () {
 70     scanf("%d", &n);
 71     scanf("%d%d", &x0, &y0);
 72     for(int i = 0; i < n; i++) {
 73         char ch;
 74         int xx, yy;
 75         getchar();
 76         scanf("%c%d%d", &ch, &xx, &yy);
 77         int dy = yy - y0, dx = xx - x0;
 78         ins(ch , dx, dy);
 79     }
 80     sort(dig1.begin(), dig1.end());
 81     sort(dig2.begin(), dig2.end());
 82     sort(dig3.begin(), dig3.end());
 83     sort(dig4.begin(), dig4.end());
 84     sort(ver1.begin(), ver1.end());
 85     sort(ver2.begin(), ver2.end());
 86     sort(ver3.begin(), ver3.end());
 87     sort(ver4.begin(), ver4.end());
 88     if(dig1.size()) {
 89         if(dig1[0].ch == 'Q' ||  dig1[0].ch == 'B') {
 90             printf("YES\n");
 91             return 0;
 92         }
 93     }
 94     if(dig2.size()) {
 95         if(dig2[0].ch == 'Q' ||  dig2[0].ch == 'B') {
 96             printf("YES\n");
 97             return 0;
 98         }
 99     }
100     if(dig3.size()) {
101         if(dig3[0].ch == 'Q' ||  dig3[0].ch == 'B') {
102             printf("YES\n");
103             return 0;
104         }
105     }
106     if(dig4.size()) {
107         if(dig4[0].ch == 'Q' ||  dig4[0].ch == 'B') {
108             printf("YES\n");
109             return 0;
110         }
111     }
112     if(ver1.size()) {
113         if(ver1[0].ch == 'Q' || ver1[0].ch == 'R') {
114             printf("YES\n");
115             return 0;
116         }
117     }
118
119     if(ver2.size()) {
120         if(ver2[0].ch == 'Q' || ver2[0].ch == 'R') {
121             printf("YES\n");
122             return 0;
123         }
124     }
125     if(ver3.size()) {
126         if(ver3[0].ch == 'Q' || ver3[0].ch == 'R') {
127             printf("YES\n");
128             return 0;
129         }
130     }
131     if(ver4.size()) {
132         if(ver4[0].ch == 'Q' || ver4[0].ch == 'R') {
133             printf("YES\n");
134             return 0;
135         }
136     }
137     printf("NO\n");
138     return 0;
139
140 }

E题：首先使用并查集将相同颜色的结点缩成一点，此时树中结点黑白相间，只需要求出树的直径（最长路）除以2就是答案。

 #include <cstdio>
 #include <cstring>
 #include <vector>
 #include <algorithm>
 #include <queue>
 using namespace std;

 const int maxn = +;
 vector<int> G[maxn];
 void add(int v, int u) {
     G[v].push_back(u);
     G[u].push_back(v);
 }
 int n;
 int col[maxn];
 int col2[maxn];
 int vis[maxn];
 int pa[maxn];
 int findpa(int x) {return x == pa[x] ? x : pa[x] = findpa(pa[x]);}
 int from, ans;
 void dfs(int u, int v, int d) {
     if(d > ans) {
         ans = d;
         from = v;
     }
     for(int i =  ; i< G[v].size(); i++) {
         int z = G[v][i];
         if(z != u) {
             dfs(v, z, d+);
         }
     }
 }
 int U[maxn];
 int V[maxn];
 int main() {
     scanf("%d", &n);
     for(int i = ; i <= n; i++)
         scanf("%d", &col[i]);

     for(int i = ; i < n; i++)
         scanf("%d%d", &U[i], &V[i]);
     for(int i = ; i <= n; i++) pa[i] = i;
     for(int i = ; i <= n; i++) {
         int fu = findpa(U[i]);
         int fv = findpa(V[i]);
         if(col[fu] == col[fv]) {
             pa[fu] = fv;
         }
     }
     for(int i = ; i <= n; i++) {
         int fu = findpa(U[i]);
         int fv = findpa(V[i]);
         if(col[fu] != col[fv]) {
             add(fu, fv);
         }
     }
     ans = -;
     int u = pa[];
     for(int i = ; i < G[u].size(); i++) {
         int v = G[u][i];
         dfs(u, v, );
     }
     u = from;
     for(int i = ; i < G[u].size(); i++) {
         int v = G[u][i];
         dfs(u, v, );
     }
     ans = (ans+)/;
     printf("%d\n", ans);
 }

F题：注意到  a&b + a|b = a+b

即可根据数组b， c的定义求出a，然后用求出的a反推数组b，c判断是否是解

详见代码：

 #include <iostream>
 #include <algorithm>
 #include <cstring>

 using namespace std;
 const int maxn =  + ;
 typedef long long LL;

 LL  a[maxn], b[maxn], c[maxn];
 int num[];
 int n;
 int main() {
     ios::sync_with_stdio(false);
     cin >> n;
     for(int i = ; i < n; i++) cin >> b[i];
     for(int i = ; i < n; i++) cin >> c[i];
     LL sum = ;
     for(int i = ; i < n; i++) {
         a[i] = b[i] + c[i];
         sum += a[i];
     }
     if(sum % (*n)) {
         cout << - << endl;
         return ;
     }
     sum /= *n;
     for(int i = ; i < n; i++) {
         a[i] = (a[i] - sum);
         if(a[i] % n) {
             cout << - << endl;
             return ;
         }
         else{
             a[i] /= n;
         }
     }
     for(int i = ; i < n; i++) {
         LL temp = a[i];
         for(int j = ; j < ; j++) {
             num[j] += temp%;
             temp /= ;
         }
     }
     for(int i = ; i < n; i++) {
         LL b_ = , c_ = ;
         for(int j = ; j < ; j++) {
             if(a[i] & (1LL << j)) {
                 b_ += num[j] * (1LL << j);
                 c_ += n * (1LL << j);
             }
             else c_ += num[j] * (1LL << j);
         }
         if(b_ != b[i] || c_ != c[i]) {
                 cout << - << endl;
                 return ;
         }
     }
     cout << a[];
     for(int i = ; i < n; i++) {
         cout << " " << a[i];
     }
     cout << endl;
     return ;
 }