POJ 3259 Wormholes bellman-ford判负环
2015-11-05 21:25
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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.
题意:给出一个图,问是否有负环。N个点,M个正边,W个负边,后面M行M行正边,双向,W条负边,单向。有负环,输出YES,没有输出NO。跑一个bellman_ford就行了。
bellman_ford原理:n个点,如果由起点到每一个点有最短路,那么由邻近的边开始类似dij的更新(他们都叫松弛,我不懂),那么剩余n-1个点,最多n-1次更新就可以找出到每个点的最短路。如果存在负环,那么这个负环可以一直跑,最短路为-inf,这样就不存在到任何点的最短路。因为已经跑了n-1次了,能最短路的都最短怒了,所以,只需要循环找一遍是否存在可以继续更新的边,如果存在,则代表存在有负环。
中间使用了一个flag,如果有点能更新就一直继续,如果不能则直接退出循环,能提高运算效率。使用结构体的储存边之间关系在这道题中效率比邻接矩阵效率高,邻接表还不会,不会讨论复杂度。而且感觉也挺好写易懂。
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.
题意:给出一个图,问是否有负环。N个点,M个正边,W个负边,后面M行M行正边,双向,W条负边,单向。有负环,输出YES,没有输出NO。跑一个bellman_ford就行了。
bellman_ford原理:n个点,如果由起点到每一个点有最短路,那么由邻近的边开始类似dij的更新(他们都叫松弛,我不懂),那么剩余n-1个点,最多n-1次更新就可以找出到每个点的最短路。如果存在负环,那么这个负环可以一直跑,最短路为-inf,这样就不存在到任何点的最短路。因为已经跑了n-1次了,能最短路的都最短怒了,所以,只需要循环找一遍是否存在可以继续更新的边,如果存在,则代表存在有负环。
中间使用了一个flag,如果有点能更新就一直继续,如果不能则直接退出循环,能提高运算效率。使用结构体的储存边之间关系在这道题中效率比邻接矩阵效率高,邻接表还不会,不会讨论复杂度。而且感觉也挺好写易懂。
/// poj3259 /// 2015/11/4 #include"stdio.h" #include"iostream" #include"algorithm" #include"string.h" #define inf 77777777; using namespace std; struct node { int u,v,w; ///u -> v 有一条边 u到v边的权值为w }edge[6010]; int n,m,w; ///n个点,m条正边,w条负边 int dis[510]; ///用来更新是否存在最短路 bool bellman_ford() { bool flag; for(int i = 1;i <= 505;i++) ///初始化 dis[i] = inf; dis[1] = 0; for(int i = 1;i <= n-1 && !flag;i++) ///n-1次更新 { flag = true; for(int j = 1;j <= m;j++) { if(dis[edge[j].v] > dis[edge[j].u]+edge[j].w) ///是否能更新 { dis[edge[j].v] = dis[edge[j].u]+edge[j].w; flag = false; ///能更新赋值为false } } } for(int i = 1;i <= m;i++) ///找是否存在能继续更新的点,有就存在负环 返回true if(dis[edge[i].v] > dis[edge[i].u]+edge[i].w) return true; return false; } int main(void) { int T; scanf("%d",&T); while(T--) { scanf("%d%d%d",&n,&m,&w); for(int i = 1;i <= m;i++) ///m条正边 { scanf("%d%d%d",&edge[i].u,&edge[i].v,&edge[i].w); edge[i+m].u = edge[i].v; ///双向边 edge[i+m].v = edge[i].u; edge[i+m].w = edge[i].w; } for(int i = m*2+1;i <= m*2+w;i++) ///w条负边 { scanf("%d%d%d",&edge[i].u,&edge[i].v,&edge[i].w); edge[i].w = -edge[i].w; } m = m*2+w; ///把m更新为总的边一共有m*2+w条 if(bellman_ford()) ///有负边输出YES,没有输出NO printf("YES\n"); else printf("NO\n"); } }
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