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POJ3259 Wormholes【BellmanFord】
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.
Lines 1..F: For each farm, output "YES" if FJ can achieve his goal, otherwise output "NO" (do not include the quotes).
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
NO
YES
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.
USACO 2006 December Gold
题目大意:农场有N个点,M条双向边,每条边信息为——从S到E花费的时间为T,同理从T到S花费的时间
也为T。现在农场出现了W个虫洞,每个虫洞意味着:从S到E花费的时间为-T(时间倒流,但是虫洞是单向边)。
问:从N个点中的某点开始走,能否使时间倒流。
思路:构建一个图,添加M条双向边,再添加W个从S到E权值为-T的负权边,求最长路径,看能否出现负权
回路,若出现负权回路,则说明能使时间倒流,否则则不能使时间倒流。
#include<iostream> #include<algorithm> #include<cstdio> #include<cstring> using namespace std; const int MAXN = 550; const int MAXM = 5500; const int INF = 0xffffff0; struct EdgeNode { int to; int w; int next; }Edges[MAXM]; int Head[MAXN],Dist[MAXN]; bool BellmanFord(int N,int M) { Dist[1] = 0; for(int i = 1; i < N; ++i) { for(int j = 1; j <= N; ++j) { if(Dist[j] == INF) continue; for(int k = Head[j]; k != -1; k = Edges[k].next) { if(Edges[k].w != INF && Dist[Edges[k].to] > Dist[j] + Edges[k].w) Dist[Edges[k].to] = Dist[j] + Edges[k].w; } } } for(int j = 1; j <= N; ++j) { if(Dist[j] == INF) continue; for(int k = Head[j]; k != -1; k = Edges[k].next) { if(Edges[k].w != INF && Dist[Edges[k].to] > Dist[j] + Edges[k].w) return true; } } return false; } int main() { int F,N,M,W,S,E,T; scanf("%d",&F); while(F--) { memset(Dist,INF,sizeof(Dist)); memset(Head,-1,sizeof(Head)); memset(Edges,0,sizeof(Edges)); scanf("%d%d%d",&N,&M,&W); int id = 0; for(int i = 0; i < M; ++i) { scanf("%d%d%d",&S,&E,&T); Edges[id].to = E; Edges[id].w = T; Edges[id].next = Head[S]; Head[S] = id++; Edges[id].to = S; Edges[id].w = T; Edges[id].next = Head[E]; Head[E] = id++; } for(int i = 0; i < W; ++i) { scanf("%d%d%d",&S,&E,&T); Edges[id].to = E; Edges[id].w = -T; Edges[id].next = Head[S]; Head[S] = id++; } if(BellmanFord(N,M)) printf("YES\n"); else printf("NO\n"); } return 0; }
POJ3259 Wormholes【BellmanFord】