| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 27964 | Accepted: 10058 |
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
解题思路：首先，应知道图不一定是连通的，只有图中存在负环就输出“YES”。因此，用spfa算法时，如果一次从源点的松弛不能到达所有边，必须把剩下的再用一次spfa算法。
#include<stdio.h>
#include<string.h>
#include<queue>
using namespace std;
#define N 505
const int inf=10005;
int map[N][N]; //记录点之间关系
int dis[N]; //记录当前点到源点的距离
int mark[N]; //标记该点是否在队列内
int num[N],n; //记录一次搜索中某点入队的次数
int Min(int a,int b)
{
return a<b?a:b;
}
int spfa()
{
int i,s=1;
queue<int>q;
memset(num,0,sizeof(num));
while(1)
{
if(q.empty())
{
for(i=1;i<=n;i++)
if(!num[i]) //若源点没有和所有的点连通，
break;
if(i<=n)
{
q.push(i); //则把后面的点入队，在进行一次spfa算法
memset(mark,0,sizeof(mark));
memset(num,0,sizeof(num));
for(;i>0;i--)
num[i]=1; //先把前面的不可到点标记
}
else
return 0;
}
s=q.front();
q.pop();
mark[s]=0;
for(i=1;i<=n;i++)
{
if(dis[i]>dis[s]+map[s][i])
{
dis[i]=dis[s]+map[s][i];
if(!mark[i])
{
mark[i]=1;
num[i]++;
q.push(i);
if(num[i]>n) //存在负环
return 1;
}
}
}
}
return 0;
}
int main()
{
int T,a,b,c,m,w,i,j;
scanf("%d",&T);
while(T--)
{
scanf("%d%d%d",&n,&m,&w);
for(i=1;i<=n;i++)
{
for(j=1;j<=n;j++)
map[i][j]=inf;
dis[i]=inf;
}
while(m--)
{
scanf("%d%d%d",&a,&b,&c);
map[a][b]=Min(map[a][b],c); //两点存在多条路
map[b][a]=Min(map[b][a],c); //bidirectional双向边
}
while(w--)
{
scanf("%d%d%d",&a,&b,&c);
map[a][b]=Min(map[a][b],0-c);
}
if(spfa())
printf("YES\n");
else
printf("NO\n");
}
return 0;
}
poj 3259 Wormholes,布布扣,bubuko.com
原文：http://blog.csdn.net/u011721440/article/details/23459893