POJ 3259 Wormholes （Bellman_Ford）

Wormholes

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 (双向)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.

POJ的题意怎么老是这么这么晦涩难懂。。。无语了都，本来英语就不咋的。。。

先说下题意吧：利用虫洞的时光旅行，如果农夫能看到过去的自己(也就是有负环)就输出YES，反之输出NO。显然要用到Bellman-Ford算法判断负环。

需要注意的是Input输入说明的部分。。很难看懂；

F:表示有多少组测试数据；

N、M、W：N个农场，M条路，W个虫洞;

S、E、T：1.S-->E所需时间(双向、S->E.E->S、正权)

  2.E-->S所需时间(单向、E->S、负权) 建图的时注意一下。

至于Bellman-Ford就不多说了。。。上代码。

#include <iostream>#include <stdio.h>#include <stdlib.h>#include <string.h>using namespace std;const int inf=99999999;int dis[inf],m,n,w,a,b,c,T;struct node{    int s,e,t;} p[inf];int Bellman (int s){    for (int i=0; i<=n; i++)    {        dis[i]=inf;    }    int flag;    for (int i=0; i<n; i++)    {        flag=0;        for (int j=0; j<s; j++)        {            if (dis[p[j].e] > dis[p[j].s]+p[j].t)            {                flag=1;               dis[p[j].e]= dis[p[j].s]+p[j].t;            }        }        if (flag==0)            break;    }    for (int j=0;j<s;j++)    {        if (dis[p[j].e] > dis[p[j].s]+p[j].t)        {            return 1;        }    }    return 0;}int main(){    scanf ("%d",&T);    while (T--)    {       int s=0;        scanf ("%d%d%d",&n,&m,&w);        for (int i=0; i<m; i++)        {            scanf ("%d%d%d",&a,&b,&c);            //双向建图            p[s].s=a;p[s].e=b;p[s++].t=c;            p[s].s=b;p[s].e=a;p[s++].t=c;        }        for (int i=0; i<w; i++)        {            scanf ("%d%d%d",&a,&b,&c);            //单向建图 && b->a            p[s].s=b;p[s].e=a;p[s++].t=-c;        }        if (Bellman(s)==0)            printf ("NO\n");        else printf ("YES\n");    }    return 0;}