Aizu - 2555 Everlasting Zero 模拟
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。
1 #pragma comment(linker, "/STACK:1000000000") 2 #include <bits/stdc++.h> 3 #define LL long long 4 #define INF 0x3f3f3f3f 5 #define IN freopen("in.txt","r",stdin); 6 #define OUT freopen("out.txt", "w", stdout); 7 using namespace std; 8 #define MAXN 105 9 #define MAXM 25005 10 11 bool dayu[MAXN][MAXN], vis[MAXN][MAXN][MAXN]; 12 int in[MAXN]; 13 queue<int> Q; 14 int gao[MAXN][MAXN], di[MAXN][MAXN]; 15 int main() 16 { 17 //IN; 18 int m, n, k, x, y; 19 char s[5]; 20 scanf("%d%d", &m, &n); 21 memset(di, 0, sizeof(di)); 22 memset(gao, INF, sizeof(gao)); 23 for(int i = 1; i <= m; i++){ 24 scanf("%d", &k); 25 for(int j = 1; j <= k; j++){ 26 scanf("%d%s%d", &x, &s, &y); 27 if(s[0] == '<'){ 28 gao[i][x] = min(gao[i][x], y); 29 } 30 else{ 31 di[i][x] = max(di[i][x], y); 32 } 33 } 34 } 35 for(int i = 1; i <= m; i++){ 36 for(int j = 1; j <= n; j++){ 37 if(di[i][j] > gao[i][j]){ 38 printf("No\n"); 39 return 0; 40 } 41 } 42 } 43 memset(dayu, 0, sizeof(dayu)); 44 memset(in, 0, sizeof(in)); 45 for(int i = 1; i <= m; i++){ 46 for(int j = 1; j <= m; j++){ 47 if(i == j) continue; 48 bool flag = false; 49 for(int k = 1; k <= n; k++){ 50 if(di[i][k] > gao[j][k]){ 51 flag = true; 52 break; 53 } 54 } 55 if(!flag){ 56 dayu[i][j] = true; 57 in[j]++; 58 } 59 } 60 } 61 for(int i = 1; i <= m; i++){ 62 for(int j = 1; j <= m; j++){ 63 for(int k = 1; k <= m; k++){ 64 if(i == j || i == k || j == k) continue; 65 if(dayu[i][j] && dayu[i][k] && dayu[k][j]){ 66 vis[i][j][k] = true; 67 vis[i][k][j] = true; 68 vis[j][i][k] = true; 69 vis[j][k][i] = true; 70 vis[k][i][j] = true; 71 vis[k][j][i] = true; 72 } 73 } 74 } 75 } 76 for(int i = 1; i <= m; i++){ 77 for(int j = 1; j <= m; j++){ 78 for(int k = 1; k <= m; k++){ 79 if(i == j || i == k || j == k) continue; 80 if(vis[i][j][k]) continue; 81 printf("No\n"); 82 return 0; 83 } 84 } 85 } 86 for(int i = 1; i <= m; i++){ 87 for(int j = 1; j <= m; j++){ 88 if(i == j) continue; 89 if(!dayu[i][j] && !dayu[j][i]){ 90 printf("No\n"); 91 return 0; 92 } 93 } 94 } 95 printf("Yes\n"); 96 return 0; 97 /* memset(vis, 0, sizeof(vis)); 98 while(!Q.empty()){ 99 Q.pop(); 100 } 101 for(int i = 1; i <= m; i++){ 102 bool flag = false; 103 for(int j = 1; j <= m; j++){ 104 if(dayu[i][j] || i == j) continue; 105 flag = true; 106 break; 107 } 108 if(!flag){ 109 Q.push(i); 110 vis[i] = true; 111 } 112 } 113 int ans = 0; 114 115 while(!Q.empty()){ 116 int s = Q.front(); 117 vis[s] = true; 118 ans++; 119 Q.pop(); 120 for(int i = 1; i <= m; i++){ 121 if(dayu[s][i]){ 122 in[i]--; 123 if(in[i] == 0 && !vis[i]){ 124 Q.push(i); 125 } 126 } 127 } 128 } 129 if(ans == m){ 130 printf("Yes\n"); 131 } 132 else{ 133 printf("No\n"); 134 }*/ 135 return 0; 136 }
