POJ3259 Wormholes 解题报告--bellman_ford

Wormholes

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 25153 Accepted: 8973

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 toF (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 toE that also moves the traveler backT 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

23 3 11 2 21 3 42 3 13 1 33 2 11 2 32 3 43 1 8

Sample Output

NOYES

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.

Source

USACO 2006 December Gold

#include<iostream>#define inf 0xfffffusing namespace std;struct s{int begin,end,time;}edge[2600];int n,m,h;bool bellman_ford(){int flag=0,i,j,k;int c=1;int dis[1001];for(i=1;i<=n;i++)dis[i]=inf;//设虚拟点到各点的距离为无穷大for(i=0;i<n;i++)//循环n-1次{flag=0;if(c++>n) break;for(j=1;j<=m;j++)//m是道路条数{if(dis[edge[j].begin]>dis[edge[j].end]+edge[j].time)//更新如0到1是否大于0到2再到1{dis[edge[j].begin]=dis[edge[j].end]+edge[j].time;flag=1;//说明有更新}if(dis[edge[j].end]>dis[edge[j].begin]+edge[j].time)//反向更新{dis[edge[j].end]=dis[edge[j].begin]+edge[j].time;flag=1;}}for(;j<=m+h;j++){if(dis[edge[j].end]>dis[edge[j].begin]-edge[j].time){dis[edge[j].end]=dis[edge[j].begin]+-edge[j].time;flag=1;}}if(flag==0) break;//如果循环一次无法更新则说明虫洞的负权不起作用，这时光倒流不可能完成}return flag;//因为这是bool函数所以返回值只有1和0，这里1说明可行//flag始终是1说明可以循环下去直到成为负环，负环即是时光倒退符合题意}int main(){int r;scanf("%d",&r);while(r--){scanf("%d%d%d",&n,&m,&h);//总数，接下来道路数，虫洞数int i,j;for(i=1;i<=m+h;i++){scanf("%d%d%d",&edge[i].begin,&edge[i].end,&edge[i].time);//把map[a][b]=c中的abc放到一个点中//因为把ab都放入一个点中，这表示a到b和b到a都是存为time的c}if(bellman_ford())//判bellprintf("YES\n");elseprintf("NO\n");}return 0;}