Wormholes(POJ 3259)

Wormholes

Time Limit:2000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

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
2 3 1
1 3 4
3 1 3
2 3 4
3 2 11 2 3
3 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在农场里探索，发现每个农场包括n块田地，m条双向路径以及w个单向虫洞。他想知道他从任意一块田地出发，能否再回到出发点。

分析：可以将题抽象成有n个顶点，m条双向边（权值为t）和w条单向边（权值为-t）。判断是否存在负环。用Bellman-Ford算法判断是否存在负环，如果不存在，则迭代n次后可求得其最短路；否则，最短路不存在（沿着负环一直走，路会更短）

<pre name="code" class="cpp">#include<stdio.h>#include<string.h>#include<iostream>#include<algorithm>using namespace std;const int MAXN = 500 + 10;const int MAXM = 5000 + 1000;const int INF = 0x3f3f3f3f;int n,m,w,k;int f[MAXN];//从顶点s指向顶点e的权值为t的边struct farm{    int s,e,t;}map[MAXM];//记录一条从顶点x到顶点y权值为z的路径void join(int x,int y,int z){    map[k].s = x;    map[k].e = y;    map[k].t = z;    k++;}//如果返回1则存在负圈，否则不存在int Bellman_Ford(){    memset(f,INF,sizeof(f));    f[1] = 0;    //求最短路    for(int i = 0;i < n;i++)    {        for(int j = 0;j < k;j++)        {            int u = map[j].s;            int v = map[j].e;            int w = map[j].t;            if(f[u] + w < f[v])                f[v] = f[u] + w;        }    }    for(int i = 0;i < n;i++)    {        for(int j = 0;j < k;j++)        {            int u = map[j].s;            int v = map[j].e;            int w = map[j].t;            if(f[u] + w < f[v])   //若还能更新，则存在负圈                return 1;        }    }    return 0;}int main(){    int t;    scanf("%d",&t);    while(t--)    {        scanf("%d %d %d",&n,&m,&w);        int x,y,z;        k = 0;        for(int i = 0;i < m;i++)        {            scanf("%d %d %d",&x,&y,&z);            join(x,y,z);            join(y,x,z);        }        for(int i = 0;i < w;i++)        {            scanf("%d %d %d",&x,&y,&z);            join(x,y,-z);        }        if(Bellman_Ford())            printf("YES\n");        else printf("NO\n");    }    return 0;}