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
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 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。
1 #include<iostream> 2 #include<cstdio> 3 using namespace std; 4 #define INF 0x3f3f3f3f 5 #define N 10100 6 int nodenum, edgenum, w, original=1; //点，边，起点 7 8 typedef struct Edge //边 9 { 10 int u; 11 int v; 12 int cost; 13 }Edge;//边的数据结构 14 15 Edge edge[N];//边 16 17 int dis[N];//距离 18 19 bool Bellman_Ford() 20 { 21 for(int i = 1; i <= nodenum; ++i) //初始化 22 dis[i] = (i == original ? 0 : INF); 23 int F=0; 24 for(int i = 1; i <= nodenum - 1; ++i)//进行nodenum-1次的松弛遍历 25 { 26 for(int j = 1; j <= edgenum*2+w; ++j) 27 { 28 if(dis[edge[j].v] > dis[edge[j].u] + edge[j].cost) //松弛（顺序一定不能反~） 29 { 30 dis[edge[j].v] = dis[edge[j].u] + edge[j].cost; 31 F=1; 32 } 33 } 34 if(!F) 35 break; 36 } 37 //与迪杰斯特拉算法类似，但不是贪心！ 38 //并没有标记数组 39 //本来松弛已经结束了 40 //但是因为由于负权环的无限松弛性 41 bool flag = 1; //判断是否含有负权回路 42 //如果存在负权环的话一定能够继续松弛 43 for(int i = 1; i <= edgenum*2+w; ++i) 44 { 45 if(dis[edge[i].v] > dis[edge[i].u] + edge[i].cost) 46 { 47 flag = 0; 48 break; 49 } 50 } 51 //只有在负权环中才能再松弛下去 52 return flag; 53 } 54 55 int main() 56 { 57 int t; 58 scanf("%d",&t); 59 while(t--) 60 { 61 62 scanf("%d %d %d", &nodenum, &edgenum, &w); 63 64 for(int i = 1; i <= 2*edgenum; i+=2)//加上道路，双向边 65 { 66 scanf("%d %d %d", &edge[i].u, &edge[i].v, &edge[i].cost); 67 edge[i+1].u=edge[i].v; 68 edge[i+1].v=edge[i].u; 69 edge[i+1].cost=edge[i].cost; 70 } 71 for(int i =2*edgenum+1; i <= 2*edgenum+w; i++)//加上虫洞，单向边，负权 72 { 73 scanf("%d %d %d", &edge[i].u, &edge[i].v, &edge[i].cost); 74 edge[i].cost=-edge[i].cost; 75 } 76 if(Bellman_Ford())//没有负环 77 printf("NO "); 78 else 79 printf("YES "); 80 } 81 return 0; 82 }