Aizu - 2555 Everlasting Zero 模拟

Aizu - 2555 Everlasting Zero

题意：学习技能，每个技能有不同的要求，问能否学习全部特殊技能

思路：枚举每两个技能，得到他们的先后学习关系，如果两个都不能先学的话就是No了，如果A>B，B>C，但是并没有A>C那么这种情况也是不允许的了，我过的也是比较惊险。

 #pragma comment(linker, "/STACK:1000000000")
 #include <bits/stdc++.h>
 #define LL long long
 #define INF 0x3f3f3f3f
 #define IN freopen("in.txt","r",stdin);
 #define OUT freopen("out.txt", "w", stdout);
 using namespace std;
 #define MAXN 105
 #define MAXM 25005

 bool dayu[MAXN][MAXN], vis[MAXN][MAXN][MAXN];
 int in[MAXN];
 queue<int> Q;
 int gao[MAXN][MAXN], di[MAXN][MAXN];
 int main()
 {
         //IN;
         int m, n, k, x, y;
         char s[];
         scanf("%d%d", &m, &n);
         memset(di, , sizeof(di));
         memset(gao, INF, sizeof(gao));
         for(int i = ; i <= m; i++){
             scanf("%d", &k);
             for(int j = ; j <= k; j++){
                 scanf("%d%s%d", &x, &s, &y);
                 if(s[] == '<'){
                     gao[i][x] = min(gao[i][x], y);
                 }
                 else{
                     di[i][x] = max(di[i][x], y);
                 }
             }
         }
         for(int i = ; i <= m; i++){
             for(int j = ; j <= n; j++){
                 if(di[i][j] > gao[i][j]){
                     printf("No\n");
                     return ;
                 }
             }
         }
         memset(dayu, , sizeof(dayu));
         memset(in, , sizeof(in));
         for(int i = ; i <= m; i++){
             for(int j = ; j <= m; j++){
                 if(i == j) continue;
                 bool flag = false;
                 for(int k = ; k <= n; k++){
                     if(di[i][k] > gao[j][k]){
                         flag = true;
                         break;
                     }
                 }
                 if(!flag){
                     dayu[i][j] = true;
                     in[j]++;
                 }
             }
         }
         for(int i = ; i <= m; i++){
             for(int j = ; j <= m; j++){
                 for(int k = ; k <= m; k++){
                     if(i == j || i == k || j == k) continue;
                     if(dayu[i][j] && dayu[i][k] && dayu[k][j]){
                         vis[i][j][k] = true;
                         vis[i][k][j] = true;
                         vis[j][i][k] = true;
                         vis[j][k][i] = true;
                         vis[k][i][j] = true;
                         vis[k][j][i] = true;
                     }
                 }
             }
         }
         for(int i = ; i <= m; i++){
             for(int j = ; j <= m; j++){
                 for(int k = ; k <= m; k++){
                     if(i == j || i == k || j == k) continue;
                     if(vis[i][j][k]) continue;
                     printf("No\n");
                     return ;
                 }
             }
         }
         for(int i = ; i <= m; i++){
             for(int j = ; j <= m; j++){
                     if(i == j) continue;
                 if(!dayu[i][j] && !dayu[j][i]){
                     printf("No\n");
                     return ;
                 }
             }
         }
         printf("Yes\n");
         return ;
        /* memset(vis, 0, sizeof(vis));
         while(!Q.empty()){
             Q.pop();
         }
         for(int i = 1; i <= m; i++){
             bool flag = false;
             for(int j = 1; j <= m; j++){
                 if(dayu[i][j] || i == j) continue;
                 flag = true;
                 break;
             }
             if(!flag){
                 Q.push(i);
                 vis[i] = true;
             }
         }
         int ans = 0;

         while(!Q.empty()){
             int s = Q.front();
             vis[s] = true;
             ans++;
             Q.pop();
             for(int i = 1; i <= m; i++){
                 if(dayu[s][i]){
                     in[i]--;
                     if(in[i] == 0 && !vis[i]){
                         Q.push(i);
                     }
                 }
             }
         }
         if(ans == m){
             printf("Yes\n");
         }
         else{
             printf("No\n");
         }*/
     return ;
 }