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Description
In the year 2163, wormholes were discovered. A wormhole is a subspace tunnel through space and time connecting two star systems. Wormholes have a few peculiar properties:
All wormholes have a constant time difference between their end points. For example, a specific wormhole may cause the person travelling through it to end up 15 years in the future. Another wormhole may cause the person to end up 42 years in the past.
A brilliant physicist, living on earth, wants to use wormholes to study the Big Bang. Since warp drive has not been invented yet, it is not possible for her to travel from one star system to another one directly. This can be done using wormholes, of
course.
The scientist wants to reach a cycle of wormholes somewhere in the universe that causes her to end up in the past. By travelling along this cycle a lot of times, the scientist is able to go back as far in time as necessary to reach the beginning of the universe
and see the Big Bang with her own eyes. Write a program to find out whether such a cycle exists.
2 3 3 0 1 1000 1 2 15 2 1 -42 4 4 0 1 10 1 2 20 2 3 30 3 0 -60
possible not possible
题意：问是否能通过虫洞，回到过去，意思实际上就是求，最短路里面有不有负环。
思路：SPFA 或者Bellman-ford 判断下有不有负环就可以了，对于SPFA，如果有负环，表明进栈次数大于等于n次。
#include <iostream>
#include <stdio.h>
#include <string>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#define INF 9999999
using namespace std;
int n,m;
int num[2222];
int dis[2222];
int vis[2222];
int f[2222];
int u[2222],v[2222],w[2222],next[2222];
int spfa()
{
queue<int>q;
memset(vis,0,sizeof vis);
memset(num,0,sizeof num);
for(int i=0;i<=n;i++) dis[i]=INF;
dis[0]=0;
q.push(0);
num[0]++;
while(!q.empty())
{
int x=q.front(); q.pop();
vis[x]=0;
for(int i=f[x];i!=-1;i=next[i])
if(dis[x]+w[i]<dis[v[i]])
{
dis[v[i]]=dis[x]+w[i];
if(vis[v[i]]==0)
{
vis[v[i]]=1;
q.push(v[i]);
num[v[i]]++;
if(num[v[i]]>=n)
return 1;
}
}
}
return 0;
}
int main()
{
int T;
scanf("%d",&T);
while(T--)
{
scanf("%d%d",&n,&m);
memset(f,-1,sizeof f);
for(int i=0;i<m;i++)
{
scanf("%d%d%d",&u[i],&v[i],&w[i]);
next[i]=f[u[i]];f[u[i]]=i;
}
if(spfa()==0)
puts("not possible");
else
puts("possible");
}
return 0;
}
/*
5 4
1 2 1
1 3 2
2 4 3
2 5 2
*/
版权声明：本文为博主原创文章，未经博主允许不得转载。
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原文地址：http://blog.csdn.net/wust_zjx/article/details/47403309
