Description
While exploring his many farms, Farmer John has discovered a number of amazing wormholes. A wormhole is very peculiar because it is a one-way path that delivers you to its destination at a time that is BEFORE you entered the wormhole! Each of FJ’s farms comprises N (1 ≤ N ≤ 500) fields conveniently numbered 1..N, M (1 ≤ M ≤ 2500) paths, and W (1 ≤ W ≤ 200) wormholes.
As FJ is an avid time-traveling fan, he wants to do the following: start at some field, travel through some paths and wormholes, and return to the starting field a time before his initial departure. Perhaps he will be able to meet himself 🙂 .
To help FJ find out whether this is possible or not, he will supply you with complete maps to F (1 ≤ F ≤ 5) of his farms. No paths will take longer than 10,000 seconds to travel and no wormhole can bring FJ back in time by more than 10,000 seconds.
Input
Line 1: A single integer, F. F farm descriptions follow. Line 1 of each farm: Three space-separated integers respectively: N, M, and W Lines 2..M+1 of each farm: Three space-separated numbers (S, E, T) that describe, respectively: a bidirectional path between S and E that requires T seconds to traverse. Two fields might be connected by more than one path. Lines M+2..M+W+1 of each farm: Three space-separated numbers (S, E, T) that describe, respectively: A one way path from S to E that also moves the traveler back T seconds.
Output
Lines 1..F: For each farm, output “YES” if FJ can achieve his goal, otherwise output “NO” (do not include the quotes).
Sample Input
2 3 3 1 1 2 2 1 3 4 2 3 1 3 1 3 3 2 1 1 2 3 2 3 4 3 1 8
Sample Output
NO YES
Hint
For farm 1, FJ cannot travel back in time. For farm 2, FJ could travel back in time by the cycle 1->2->3->1, arriving back at his starting location 1 second before he leaves. He could start from anywhere on the cycle to accomplish this.
题目大意
输入样例数，然后输入三个数“点数，双向边数，单向边数”然后分别输入双向边，单向边(权是负的)。问图中有没有负圈
解题思路
陷阱1：用algorithm中min函数会超时，手写一个就可以过了 陷阱2：题目中有负圈，，在输入边的时候要判断一下，保存最小的边
代码#include <stdio.h>const int INF = 10000000;const int maxn = 510;int eg[maxn][maxn];int n;int main(){int t;scanf(“%d”,&t);while(t–) {for(int i = 0 ; i < maxn ; i ++) {for(int j = 0 ; j < maxn ; j ++) {eg[i][j] = INF;}}int e1,e2;scanf(“,&n,&e1,&e2);while(e1–) {int a,b,c;scanf(“,&a,&b,&c);if(eg[a][b] < c) continue;eg[a][b] = c;eg[b][a] = c;}while(e2–) {int a,b,c;scanf(“,&a,&b,&c);eg[a][b] = -c;}int isthere = 0;for(int k = 1 ; k <= n ; k ++) {for(int i = 1 ; i <= n ; i ++) {for(int j = 1 ; j <= n ; j ++) {if(eg[i][k]+eg[k][j] < eg[i][j]) eg[i][j] = eg[i][k]+eg[k][j];}if(eg[i][i] < 0) {isthere = 1;break;}}if(isthere) break;}if(!isthere) printf(“NO\n”);else printf(“YES\n”);}return 0;}
而是自己。在你成功地把自己推销给别人之前，
