POJ-3259-Wormholes

Wormholes

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 43757 Accepted: 16066

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

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.

#include<stdio.h>#include<string.h>#include<queue>#include<vector>using namespace std;const int MAX = 501;const int INF = 9999999;int t, n,m,w, dis[MAX];struct node{    int to;    int w;}ver;vector<node>graph[MAX];bool vis[MAX];void SPFA(int s){    queue<int>qq;    memset(vis,false,sizeof(vis));    for(int i = 1;i<=n;++i)        dis[i] = INF;    dis[s] = 1;    qq.push(s);    while(!qq.empty())    {        int f = qq.front();        qq.pop();        vis[f] = false;        for(int i= 0;i<(int)graph[f].size();++i)        {            int to = graph[f][i].to;            if(dis[f]<INF&&dis[f]+graph[f][i].w<dis[to])            {                dis[to] = dis[f]+graph[f][i].w;                if(to==1&&dis[to]<0)return;// 如果dis[1]<0说明可以回到过去                if(!vis[to])                {                    qq.push(to);                    vis[to] = true;                }            }        }    }}int main(){    int a,b,c;    scanf("%d", &t);    while(t--)    {        scanf("%d%d%d",&n,&m,&w);        for(int i = 0; i<=n; ++i)            graph[i].clear();        while(m--)        {            scanf("%d%d%d", &a, &b,&c);            ver.to = b;            ver.w = c;            graph[a].push_back(ver);            ver.to = a;            ver.w = c;            graph[b].push_back(ver);        }        while(w--)        {            scanf("%d%d%d",&a,&b,&c);            ver.to = b;            ver.w = -c;            graph[a].push_back(ver);        }        SPFA(1);        if(dis[1]<0)printf("YES\n");        else printf("NO\n");    }    return 0;}