Wormholes(最短路_bellman_ford)
2014-10-22 14:13
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Wormholes
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
Sample Output
Time Limit: 2000MS | Memory Limit: 65536K | |
Total Submissions: 31762 | Accepted: 11561 |
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
YESHintFor 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:John的农场里N块地,M条路连接两块地,W个虫洞,虫洞是一条单向路,会在你离开之前把你传送到目的地,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。思路:其实就是看看有没有负环,如果有负环的话就证明能回去就输出YES,没有就输出NO。可以用贝尔曼福德判断一下负环 #include <stdio.h> #include <string.h> #include <stdlib.h> #include <algorithm> using namespace std; #define inf 0x3f3f3f3f struct node { int u,v,w; } edge[6010]; int dis[510]; int cnt; int n,m,W; void add_edge(int u,int v,int w) { edge[cnt].u=u; edge[cnt].v=v; edge[cnt].w=w; cnt++; } int bellman_ford() { int i,j; for(i=1; i<=n; i++) dis[i]=inf; dis[1]=0; for(i=1; i<n; i++) { int flag=0; for(j=0; j<cnt; j++) { if(dis[edge[j].v]>dis[edge[j].u]+edge[j].w) { dis[edge[j].v]=dis[edge[j].u]+edge[j].w; flag=1; } } if(!flag) break; } for(i=0; i<cnt; i++) if(dis[edge[i].v]>dis[edge[i].u]+edge[i].w) return 1; return 0; } int main() { int T; int u,v,w; scanf("%d",&T); while(T--) { cnt=0; scanf("%d %d %d",&n,&m,&W); while(m--) { scanf("%d %d %d",&u,&v,&w); add_edge(u,v,w); add_edge(v,u,w); } while(W--) { scanf("%d %d %d",&u,&v,&w); add_edge(u,v,-w); } if(bellman_ford()) printf("YES\n"); else printf("NO\n"); } return 0; }
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