Aizu

 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[5];
        scanf("%d%d", &m, &n);
        memset(di, 0, sizeof(di));
        memset(gao, INF, sizeof(gao));
        for(int i = 1; i <= m; i++){
            scanf("%d", &k);
            for(int j = 1; j <= k; j++){
                scanf("%d%s%d", &x, &s, &y);
                if(s[0] == '<'){
                    gao[i][x] = min(gao[i][x], y);
                }
                else{
                    di[i][x] = max(di[i][x], y);
                }
            }
        }
        for(int i = 1; i <= m; i++){
            for(int j = 1; j <= n; j++){
                if(di[i][j] > gao[i][j]){
                    printf("No
");
                    return 0;
                }
            }
        }
        memset(dayu, 0, sizeof(dayu));
        memset(in, 0, sizeof(in));
        for(int i = 1; i <= m; i++){
            for(int j = 1; j <= m; j++){
                if(i == j) continue;
                bool flag = false;
                for(int k = 1; k <= n; k++){
                    if(di[i][k] > gao[j][k]){
                        flag = true;
                        break;
                    }
                }
                if(!flag){
                    dayu[i][j] = true;
                    in[j]++;
                }
            }
        }
        for(int i = 1; i <= m; i++){
            for(int j = 1; j <= m; j++){
                for(int k = 1; 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 = 1; i <= m; i++){
            for(int j = 1; j <= m; j++){
                for(int k = 1; k <= m; k++){
                    if(i == j || i == k || j == k) continue;
                    if(vis[i][j][k]) continue;
                    printf("No
");
                    return 0;
                }
            }
        }
        for(int i = 1; i <= m; i++){
            for(int j = 1; j <= m; j++){
                    if(i == j) continue;
                if(!dayu[i][j] && !dayu[j][i]){
                    printf("No
");
                    return 0;
                }
            }
        }
        printf("Yes
");
        return 0;
       /* 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
");
        }
        else{
            printf("No
");
        }*/
    return 0;
}