Bellman_ford POJ 3259 Wormholes
| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 41728 | Accepted: 15325 |
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.
就是当你过去的时候时间会倒退Ts。我们的任务是知道会不会在从某块地出发后又回来，看到了离开之前的自己。
简化下，就是看图中有没有负权环。
1 #include <cstdio> 2 #define maxn 502 3 #define INF 0x3fffffff 4 5 ///边 6 typedef struct Edge{ 7 int u, v; ///起点， 终点 8 int weight; ///权值 9 }Edge; 10 11 Edge edge[6000]; ///双向边，保存边的权值 12 int dist[maxn]; ///节点到原点的最小距离 13 int edgenum; ///边数 14 ///插入边 15 void insert(int u, int v, int w) 16 { 17 edge[edgenum].u = u; 18 edge[edgenum].v = v; 19 edge[edgenum++].weight = w; 20 } 21 bool Bellman_Ford(int source, int nodenum) ///原点和结点个数 22 { 23 for(int i=0; i < nodenum; ++i) 24 dist[i] = INF; 25 dist[source] = 0; 26 for(int i=1; i < nodenum; ++i){ 27 for(int j=0; j < edgenum; ++j) 28 { 29 if(dist[edge[j].v] > dist[edge[j].u] + edge[j].weight)///松弛计算 30 dist[edge[j].v] = dist[edge[j].u] + edge[j].weight; 31 } 32 } 33 bool flag = false; 34 /// 判断是否有负权环 35 for(int i=0; i < edgenum; ++i){ 36 if(dist[edge[i].v] > dist[edge[i].u] + edge[i].weight) 37 { 38 flag = true; ///有负权环 39 break; 40 } 41 } 42 return flag; 43 } 44 int main() 45 { 46 int T; 47 scanf("%d", &T); 48 while(T--) 49 { 50 int n, m, c; 51 scanf("%d%d%d", &n, &m, &c); 52 edgenum = 0; 53 for(int i=0; i<m; i++) 54 { 55 int u, v, w; 56 scanf("%d%d%d", &u, &v, &w); 57 insert(u, v, w); 58 insert(v, u, w); 59 } 60 for(int i=0; i<c; i++) 61 { 62 int u, v, w; 63 scanf("%d%d%d", &u, &v, &w); 64 insert(u, v, -w); 65 } 66 if(Bellman_Ford(0, n)) 67 printf("YES\n"); 68 else 69 printf("NO\n"); 70 } 71 return 0; 72 }
