poj 3259 Wormholes[ bellman_ford 判负环]

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 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 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");        else            printf("NO\n");    }    return 0;}