http://poj.org/problem?id=3259
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 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
Sample Input
Sample Output
NO
YES
Hint
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.
题目大意：
时空旅行，前m条路是双向的，旅行时间为正值，w条路是虫洞，单向的，旅行时间是负值，也就是能回到过去。求从一点出发，判断能否在”过去“回到出发点，即会到出发点的时间是负的。
解题思路：
裸的负权最短路问题，SPFA Bellman-Ford解决。
#include<iostream>
#include<cstdio>
using namespace std;
#define INF 0x3f3f3f3f
#define N 10100
int nodenum, edgenum, w, original=; //点，边，起点 typedef struct Edge //边
{
int u;
int v;
int cost;
}Edge;//边的数据结构 Edge edge[N];//边 int dis[N];//距离 bool Bellman_Ford()
{
for(int i = ; i <= nodenum; ++i) //初始化
dis[i] = (i == original ? : INF);
int F=;
for(int i = ; i <= nodenum - ; ++i)//进行nodenum-1次的松弛遍历
{
for(int j = ; j <= edgenum*+w; ++j)
{
if(dis[edge[j].v] > dis[edge[j].u] + edge[j].cost) //松弛（顺序一定不能反~）
{
dis[edge[j].v] = dis[edge[j].u] + edge[j].cost;
F=;
}
}
if(!F)
break;
}
//与迪杰斯特拉算法类似，但不是贪心！
//并没有标记数组
//本来松弛已经结束了
//但是因为由于负权环的无限松弛性
bool flag = ; //判断是否含有负权回路
//如果存在负权环的话一定能够继续松弛
for(int i = ; i <= edgenum*+w; ++i)
{
if(dis[edge[i].v] > dis[edge[i].u] + edge[i].cost)
{
flag = ;
break;
}
}
//只有在负权环中才能再松弛下去
return flag;
} int main()
{
int t;
scanf("%d",&t);
while(t--)
{ scanf("%d %d %d", &nodenum, &edgenum, &w); for(int i = ; i <= *edgenum; i+=)//加上道路，双向边
{
scanf("%d %d %d", &edge[i].u, &edge[i].v, &edge[i].cost);
edge[i+].u=edge[i].v;
edge[i+].v=edge[i].u;
edge[i+].cost=edge[i].cost;
}
for(int i =*edgenum+; i <= *edgenum+w; i++)//加上虫洞，单向边，负权
{
scanf("%d %d %d", &edge[i].u, &edge[i].v, &edge[i].cost);
edge[i].cost=-edge[i].cost;
}
if(Bellman_Ford())//没有负环
printf("NO\n");
else
printf("YES\n");
}
return ;
}