Wormholes(最短路_bellman_ford)

Wormholes

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 31762 Accepted: 11561

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

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.
题意：John的农场里N块地，M条路连接两块地，W个虫洞，虫洞是一条单向路，会在你离开之前把你传送到目的地，就是当你过去的时候时间会倒退Ts。我们的任务是知道会不会在从某块地出发后又回来，看到了离开之前的自己。
思路：其实就是看看有没有负环，如果有负环的话就证明能回去就输出YES，没有就输出NO。可以用贝尔曼福德判断一下负环
#include <stdio.h>#include <string.h>#include <stdlib.h>#include <algorithm>using namespace std;#define inf 0x3f3f3f3fstruct node{    int u,v,w;} edge[6010];int dis[510];int cnt;int n,m,W;void add_edge(int u,int v,int w){    edge[cnt].u=u;    edge[cnt].v=v;    edge[cnt].w=w;    cnt++;}int bellman_ford(){    int i,j;    for(i=1; i<=n; i++)        dis[i]=inf;    dis[1]=0;    for(i=1; i<n; i++)    {        int flag=0;        for(j=0; j<cnt; j++)        {            if(dis[edge[j].v]>dis[edge[j].u]+edge[j].w)            {                dis[edge[j].v]=dis[edge[j].u]+edge[j].w;                flag=1;            }        }        if(!flag)            break;    }    for(i=0; i<cnt; i++)        if(dis[edge[i].v]>dis[edge[i].u]+edge[i].w)            return 1;    return 0;}int main(){    int T;    int u,v,w;    scanf("%d",&T);    while(T--)    {        cnt=0;        scanf("%d %d %d",&n,&m,&W);        while(m--)        {            scanf("%d %d %d",&u,&v,&w);            add_edge(u,v,w);            add_edge(v,u,w);        }        while(W--)        {            scanf("%d %d %d",&u,&v,&w);            add_edge(u,v,-w);        }        if(bellman_ford())            printf("YES\n");        else            printf("NO\n");    }    return 0;}