[ACM] POJ 3259 Wormholes (bellman-ford最短路径，推断是否存在负权回路）

Wormholes

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 29971		Accepted: 10844

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

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.

Source

USACO 2006 December Gold

解题思路：

这题什么意思读了半天也没弄懂。。在POJ上做题头一大难关就是读题意。。。

农民有N块地，每块地看做一个节点，有m条普通的路（双向），连接着两个节点，从一端走到还有一端须要w时间（权值），还有wh条特殊的单向路，也就是题意中的虫洞，虫洞也连接着两个节点，但虫洞是单向的，从起点走到终点须要w时间，但这个时间是负的，也就是题意中所说的时光倒流，比方 一个虫洞连接着s -> e，在s处如果时间为6，走虫洞时间为4，那么走过去时光倒流，走到e时时间就变为了2 （6 -4，也就是虫洞这条路的权值为 -4 ) ，好奇妙。。。。问有没有这样一种情况，就是第二次走到某个节点的时间比第一次走到该节点所用的时间短（题中的Perhaps
he will be able to meet himself，时光倒流的作用）。如果存在这样的情况，那么以某节点为起点和终点一定存在着一个回路，这个回路的权值是负的，这样第二次所用的时间一定比第一次少。

bellman-ford求最短路径算法中的第三步就是推断一个图（有向图，或无向图）中是否存在负权回路。

bellman-ford算法參考：http://blog.csdn.net/niushuai666/article/details/6791765

代码：

#include <iostream>
#include <iostream>
using namespace std;
const int inf=10010;
const int maxn=6000;

struct Edge//边结构体
{
    int s,e,w;
}edge[maxn];

int dis[maxn];//到达各顶点的距离
int nodeNum,edgeNum;//节点个数，边个数

bool bellman_ford()
{
    for(int i=0;i<=nodeNum;i++)
        dis[i]=inf;//初始化
    bool ok;//推断是否发生了松弛
    for(int i=1;i<=nodeNum-1;i++)
    {
        ok=0;
        for(int j=1;j<=edgeNum;j++)
        {
            if(dis[edge[j].s]+edge[j].w<dis[edge[j].e])
            {
                dis[edge[j].e]=dis[edge[j].s]+edge[j].w;
                ok=1;
            }
        }
        if(!ok)//没有发生松弛，及时退出
            break;
    }
    for(int i=1;i<=edgeNum;i++)//寻找负权回路
        if(dis[edge[i].s]+edge[i].w<dis[edge[i].e])
            return true;//存在负权回路
    return false;
}

int main()
{
    int t;cin>>t;
    int n,m,wh;
    while(t--)
    {
        cin>>n>>m>>wh;//n为节点，m为双向边，wh为单向边
        nodeNum=n;
        edgeNum=m*2+wh;
        int cnt=1;
        int s,e,w;//起点，终点，权值
        for(int i=1;i<=m;i++)
        {
            cin>>s>>e>>w;
            edge[cnt].s=s;
            edge[cnt].e=e;
            edge[cnt++].w=w;
            edge[cnt].s=e;
            edge[cnt].e=s;
            edge[cnt++].w=w;
        }
        for(int i=1;i<=wh;i++)
        {
            cin>>s>>e>>w;
            edge[cnt].s=s;
            edge[cnt].e=e;
            edge[cnt++].w=-w;//注意是负权
        }
        if(bellman_ford())//存在负权回路
            cout<<"YES"<<endl;
        else
            cout<<"NO"<<endl;
    }
    return 0;
}