POJ

问题描述
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

分析：

代码如下：

#include<cstdio>#include<cstring>using namespace std;const int maxn = 500+10; const int maxe = 5200+10;struct edge{    int from, to, cost;}es[maxe];int F;int V,E;int d[maxn];/*判断是否有负圈*/bool find_negative_loop(){    memset(d, 0, sizeof(d));    for(int i=0; i<V; i++)    {        for(int j=0; j<E; j++)        {            edge e = es[j];            if(d[e.to] > d[e.from] + e.cost)            {                d[e.to] = d[e.from] + e.cost;                if(i == V-1) return true;            }        }    }    return false;}int main(){    scanf("%d",&F);    while(F--)    {        int N,M,W;        scanf("%d%d%d",&N,&M,&W);        V = N;        E = 0;         int s,e,t;        for(int i=0; i<M; i++)        {            scanf("%d%d%d",&s,&e,&t);            es[E].from = s;            es[E].to = e;            es[E].cost = t;            E++;            es[E].to = s;            es[E].from = e;            es[E].cost = t;            E++;        }           for(int i=0; i<W; i++)        {            scanf("%d%d%d",&s,&e,&t);            es[E].from = s;            es[E].to = e;            es[E].cost = -t;            E++;        }        if(find_negative_loop()) printf("YES\n");        else printf("NO\n");    }    return 0;}