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| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 35103 | Accepted: 12805 |
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
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
Source
题目大意：虫洞问题，现在有n个点，m条边，代表现在可以走的通路，比如从a到b和从b到a需要花费c时间，现在在地上出现了w个虫洞，虫洞的意义就是你从a到b话费的时间是-c(时间倒流,并且虫洞是单向的)，现在问你从某个点开始走，能回到从前
解题思路：其实给出了坐标，这个时候就可以构成一张图，然后将回到从前理解为是否会出现负权环，用bellman-ford就可以解出了
#include<stdio.h>
#include<string.h>
#include<stack>
#include<iostream>
#include<algorithm>
using namespace std;
struct node{
int u,v,w;
}que[5400];
int n,m,wh;
int Count;
int inf=999999999;
int dis[5000];
bool bellman_ford(){
memset(dis,inf,sizeof(dis));
dis[1]=0;
int flag;
int a,b,c;
for(int i=1;i<n;i++){
flag=0;
for(int j=0;j<Count;j++){
a=que[j].u,b=que[j].v,c=que[j].w;
if(dis[b]>dis[a]+c){
dis[b]=dis[a]+c;
flag=1;
}
}
if(!flag)
break;
}
for(int j=0;j<Count;j++){
a=que[j].u,b=que[j].v,c=que[j].w;
if(dis[b]>dis[a]+c)
return true;}
return false;
}
int main(){
int t;
scanf("%d",&t);
while(t--){
Count=0;
scanf("%d%d%d",&n,&m,&wh);
int t1,t2,t3;
for(int i=1;i<=m;i++){
scanf("%d%d%d",&t1,&t2,&t3);
que[Count].u=t1;
que[Count].v=t2;
que[Count].w=t3;
Count++;
que[Count].u=t2;
que[Count].v=t1;
que[Count].w=t3;
Count++;
}
for(int i=m+1;i<=m+wh;i++){
scanf("%d%d%d",&t1,&t2,&t3);
que[Count].u=t1;
que[Count].v=t2;
que[Count].w=-t3;
Count++;
}
if(bellman_ford())
printf("YES\n");
else
printf("NO\n");
}
return 0;
}
poj3259 bellman——ford Wormholes解绝负权问题
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原文地址：http://www.cnblogs.com/13224ACMer/p/4634618.html
