obsidian

POJ 3259 Wormholes (Bellman-Ford判断负环)

07
2017/08

# 题目

## 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 $N (1 \leq N \leq 500)$ fields conveniently numbered $1...N$, $M$ $(1 \leq M \leq 2500)$ paths, and $W$ $(1 \leq W \leq 200)$ 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 $F$ $(1 \leq F \leq 5)$ 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
YES

## Hint

For farm $1$, FJ cannot travel back in time.

For farm $2$ , FJ could travel back in time by the cycle $1\rightarrow2\rightarrow3\rightarrow1$ , arriving back at his starting location $1$ second before he leaves. He could start from anywhere on the cycle to accomplish this.

# 代码

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<algorithm>
#include<queue>
#include<stack>
#include<vector>
#include<cmath>
#include<set>
#include<map>
#include<cstdlib>
#include<functional>
#include<climits>
#include<cctype>
#include<iomanip>
using namespace std;
typedef long long ll;
#define INF 0x3f3f3f3f
#define mod 1e9+7
#define clr(a,x) memset(a,x,sizeof(a))
const double eps = 1e-6;
const int MAX_N=10000;
const int MAX_E=10000;
int From[MAX_E],To[MAX_E],W[MAX_E];
int dis[MAX_N],tot;
void init()
{
tot=0;
}
{
From[tot]=u;
To[tot]=v;
W[tot++]=d;
}
bool Bellman_ford(int s,int n)
{
clr(dis,0x3f);
dis[s]=0;
for(int k=0;k<n-1;k++)
{
bool relaxed=0;
for(int i=0;i<tot;i++)
{
int x=From[i],y=To[i];
if(dis[y]>dis[x]+W[i])
{
dis[y]=dis[x]+W[i];
relaxed=1;
}
}
if(!relaxed)
break;
}
for(int i=0;i<tot;i++)
if(dis[To[i]]>dis[From[i]]+W[i])
return 1;
return 0;
}
int main()
{
int f,n,m,t,u,v,w;
cin>>f;
while(f--)
{
int num=0;
init();
cin>>n>>m>>t;
//*************************************建图
for(int i=0;i<m;i++)
{
cin>>u>>v>>w;
num+=2; //记录边的个数
}
for(int i=0;i<t;i++)
{
cin>>u>>v>>w;
}
//*************************************建图
bool ans=Bellman_ford(1,num);
if(ans)
cout<<"YES"<<endl;
else
cout<<"NO"<<endl;
}
return 0;
}

Last modification：November 2nd, 2019 at 03:02 pm