| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 30766 | Accepted: 11157 |
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
题意：有双向的正边和单项的负边（虫洞的时间是倒流的），问是否能走回去。
思路：用bellman_ford算法判断是否存在负权环。
#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cmath>
#include <cstring>
#include <stdlib.h>
using namespace std;
const int INF=1000000009;
struct Node{
int s,e,t;
}mp[5205];
int d[505];
int n,m,w;
bool bellman_ford(){
memset(d,INF,sizeof(d));
d[1]=0;
for(int i=1;i<n;i++){
bool flag=true;
for(int j=1;j<=2*m+w;j++){
int a=mp[j].s,b=mp[j].e,c=mp[j].t;
if(d[b]>d[a]+c){
d[b]=d[a]+c;
flag=false;
}
}
if(flag) false;
}
for(int i=1;i<=2*m+w;i++){
int a=mp[i].s,b=mp[i].e,c=mp[i].t;
if(d[b]>d[a]+c)
return false;
}
return true;
}
int main(){
int t;
scanf("%d",&t);
while(t--){
memset(mp,0,sizeof(mp));
scanf("%d%d%d",&n,&m,&w);
int k=0;
for(int i=1;i<=m;i++){
int x,y,z;
scanf("%d%d%d",&x,&y,&z);
++k;
mp[k].s=x,mp[k].e=y,mp[k].t=z;
++k;
mp[k].e=x,mp[k].s=y,mp[k].t=z;
}
for(int i=m+1;i<=m+w;i++){
int x,y,z;
scanf("%d%d%d",&x,&y,&z);
++k;
mp[k].s=x,mp[k].e=y,mp[k].t=-z;
}
if(!bellman_ford()) printf("YES\n");
else printf("NO\n");
}
return 0;
}
poj 3259 Wormholes,布布扣,bubuko.com
原文：http://blog.csdn.net/kimi_r_17/article/details/38349117