poj 3259 Wormholes

Wormholes

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 29412 Accepted: 10631

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.

一个最短路的问题，但因为有负值，需用Bellman-Ford

#include <iostream>#include <stdio.h>#include<stdlib.h>#include<string.h>#define MAX 999990using namespace std;struct node{    int s,e,w;}q[10000];int n,m,w,r;int v[10000];void bellman(){    int i,j;    for(i=1;i<=n-1;i++)                 //松弛    {        for(j=0;j<r;j++)                        {           if(v[q[j].e]>v[q[j].s]+q[j].w)           {               v[q[j].e]=v[q[j].s]+q[j].w;                         }        }    }    int f=1;     for(j=0;j<r;j++)                   //判断是否有回路     {         if(v[q[j].e]>v[q[j].s]+q[j].w)           {               f=0;               break;           }     }     if(f)        printf("NO\n");     else printf("YES\n");}int main(){    int s,e,t,y;    scanf("%d",&y);    while(y--)    {        scanf("%d%d%d",&n,&m,&w);       for(int i=0;i<=n;i++)        v[i]=MAX;        v[1]=0;       r=0;       while(m--)                    //双向        {            scanf("%d%d%d",&s,&e,&t);            q[r].s=s;            q[r].e=e;            q[r].w=t;            r++;            q[r].s=e;            q[r].e=s;            q[r].w=t;            r++;        }            while(w--)               //单向  存在负值             {                 scanf("%d%d%d",&s,&e,&t);                 q[r].s=s;                 q[r].e=e;                 q[r].w=-t;                 r++;             }             bellman();    }    return 0;}