HDU 3259 Wormholes（Bellman-Ford算法）

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

2
3 3 1
1 2 2
1 3 4
2 3 1
3 1 3
3 2 1
1 2 3
2 3 4
3 1 8

Sample Output

NO
YES

题目大意

第一行 输入一个数 是表示几组测试数据

第二行 三个数 N(点的个数),M(正边的个数),W(负边的个数) 注意 ：正边为双向的，负边为单向的。然后 M行u,v,w，再然后W行u,v,w。求这个图是不是存在负环。 有 YES 没NO。

参考代码

#include <iostream>#include <cstdio>#define INF 0x3f3f3fusing namespace std;int dis[1005];int n, m, k, t;struct node{    int v, u, w;}a[5005];bool bellmanFord(){    for (int i = 1;i <= n - 1;i++)    {        int flag = 0;        for (int j = 1;j <= m;j++)        {            if (dis[a[j].u] > dis[a[j].v] + a[j].w)            {                dis[a[j].u] = dis[a[j].v] + a[j].w;                flag = 1;            }        }        if (!flag)            break;    }    for (int j = 1;j <= m;j++)    {           if (dis[a[j].u] > dis[a[j].v] + a[j].w)        {            return true;        }    }    return false;}int main(){    scanf("%d", &t);    while (t--)    {        scanf("%d %d %d", &n, &m, &k);        for (int i = 1;i <= n;i++)            dis[i] = INF;        for (int i = 1;i <= m;i++)        {            scanf("%d %d %d", &a[i].u, &a[i].v, &a[i].w);            a[i + m].v = a[i].u;            a[i + m].u = a[i].v;            a[i + m].w = a[i].w;        }        m = 2 * m + 1;        for (int i = m;i < m + k;i++)        {            scanf("%d %d %d", &a[i].u, &a[i].v, &a[i].w);            a[i].w = -a[i].w;        }        m += k;        dis[1] = 0;        if (bellmanFord())            printf("YES\n");        else            printf("NO\n");    }}