Bellman-Ford算法模板题——POJ 3259
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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 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
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
懒得自己打题意~
这道题是是求是否存在负权环。
题意 ： 一个famer有一些农场，这些农场里面有一些田地，田地里面有一些虫洞，田地和田地之间有路，虫洞有这样的性质： 时间倒流。问你这个农民能不能看到他自己，也就是说，有没有这样一条路径，能利用虫洞的时间倒流的性质，让这个人能在这个点出发前回去，这样他就是能看到他自己
解题思路：使用Bellman-Ford算法，看图中有没有负权环。有的话就是可以，没有的话就是不可以了。
输入N,M,W，表示N个点，M表示两点间无向，W表示两点间有向
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <cstring>
#include <queue>
#include <algorithm>
using namespace std;
const int MAXN=10010;
const int INF=1<<30;
int dis[505];
struct edge
{
int u,v,t;
}E[6000];
bool Bellman_Ford(int n, int m)
{
int i,j;
for(i=1; i<=n; i++) dis[i]=INF;
dis[1]=0;
for(i=1; i<=n-1; i++){
for(j=0; j<m; j++){
if(dis[E[j].u]+E[j].t < dis[E[j].v]) dis[E[j].v] = dis[E[j].u]+E[j].t;
//cout<<E[j].u<<" "<<E[j].v<<" "<<dis[E[j].u]<<" "<<dis[E[j].v]<<endl;
}
}
for(j=0; j<m; j++){
if(dis[E[j].v] > dis[E[j].u]+E[j].t) return 1;
}
return 0;
}
int main()
{
//freopen("in.txt","r",stdin);
int T;
scanf("%d", &T);
while(T--)
{
int n,m,w;
scanf("%d%d%d", &n,&m,&w);
int size=0;
int i,j;
for(i=1; i<=m; i++){
int u,v,t;
scanf("%d%d%d", &u,&v,&t);
E[size].u=u;
E[size].v=v;
E[size].t=t;
size++;
E[size].v=u;
E[size].u=v;
E[size].t=t;
size++;
}
for(i=1; i<=w; i++){
int u,v,t;
scanf("%d%d%d", &u,&v,&t);
E[size].u=u;
E[size].v=v;
E[size].t=-t;
size++;
}
int ok=Bellman_Ford(n,size);
if(ok) printf("YES\n");
else printf("NO\n");
//cout<<m+w<<" "<<size<<endl;
}
return 0;
}