Wormholes
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 toF (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 WLines 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 toE 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
23 3 11 2 21 3 42 3 13 1 33 2 11 2 32 3 43 1 8
Sample Output
NOYES
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.
思路：普通路双向，正值；虫洞单向，负值；bellman_ford 判负环；
代码：
#include<iostream>#include<cstdio>#include<cstring>#include<cmath>#include<vector>#include<queue>#include<stack>#include<map>#define INF 0x3f3f3f3f#define mem(a,b) memset(a,b,sizeof(a))using namespace std;typedef long long LL;const int N=1002;int dist[N];int gra[N][N],vis[N][N];bool bellman_ford(int n,int s){for(int i=1;i<=n;i++){dist[i]=INF;}dist[s]=0;for(int i=1;i<=n-1;i++){int flag=false;for(int j=1;j<=n;j++){for(int k=1;k<=n;k++){if(vis[j][k]==1&&dist[k]>dist[j]+gra[j][k]){dist[k]=dist[j]+gra[j][k];flag=true;}}}if(!flag) return false;}for(int j=1;j<=n;j++){for(int k=1;k<=n;k++){if(vis[j][k]==1&&dist[k]>dist[j]+gra[j][k])return true;}}return false;}int main(){int T;scanf("%d",&T);while(T–){int n,m,w;scanf("%d%d%d",&n,&m,&w);mem(vis,0);int s,e,t;for(int i=0;i<m;i++){scanf("%d%d%d",&s,&e,&t);if(vis[s][e]){if(gra[s][e]>t)gra[s][e]=gra[e][s]=t;}else{gra[s][e]=gra[e][s]=t;vis[s][e]=vis[e][s]=1;}}for(int i=0;i<w;i++){scanf("%d%d%d",&s,&e,&t);gra[s][e]=0-t;vis[s][e]=1;}int flag=0;for(int i=1;i<=1;i++)//只用判1，，就行了，图是联通的，不然临接矩阵超时！！{if(bellman_ford(n,i)){flag=1;break;}}//for(int i=1;i<=n;i++)//printf("%d___",dist[i]);//printf("%d\n",gra[3][1]);if(flag)printf("YES\n");elseprintf("NO\n");}return 0;}
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