| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 12913 | Accepted: 5778 |
Description
One cow from each of N farms (1 ≤ N ≤ 1000) conveniently numbered 1..N is going to attend the big cow party to be held at farm #X (1 ≤ X ≤ N). A total of M (1 ≤ M ≤ 100,000) unidirectional (one-way roads connects pairs of farms; road i requires Ti (1 ≤ Ti ≤ 100) units of time to traverse.
Each cow must walk to the party and, when the party is over, return to her farm. Each cow is lazy and thus picks an optimal route with the shortest time. A cow‘s return route might be different from her original route to the party since roads are one-way.
Of all the cows, what is the longest amount of time a cow must spend walking to the party and back?
Input
Output
Sample Input
4 8 2 1 2 4 1 3 2 1 4 7 2 1 1 2 3 5 3 1 2 3 4 4 4 2 3
Sample Output
10
Hint
思路：用两个二维数组分别存边的信息，两次dijkstra求得最短路。也可以用spfa求，速度很慢。
#include"stdio.h"
#include"string.h"
#define N 1005
const int inf=0x7fffffff;
int n,m,x;
int g1[N][N],g2[N][N];
int dis1[N],dis2[N],mark[N];
void dijkstra(int g[N][N],int *dis)
{
int i,u,min;
for(i=1;i<=n;i++)
{
dis[i]=inf;
mark[i]=0;
}
dis[x]=0;
while(1)
{
u=0;
min=inf;
for(i=1;i<=n;i++)
{
if(min>dis[i]&&!mark[i])
{
min=dis[i];
u=i;
}
}
if(u==0)
break;
mark[u]=1;
for(i=1;i<=n;i++)
{
if(g[u][i]<inf&&dis[i]>dis[u]+g[u][i])
dis[i]=dis[u]+g[u][i];
}
}
}
int main()
{
int i,j,u,v,w;
while(scanf("%d%d%d",&n,&m,&x)!=-1)
{
for(i=1;i<=n;i++)
{
for(j=1;j<=n;j++)
g1[i][j]=g2[i][j]=inf;
}
for(i=0;i<m;i++)
{
scanf("%d%d%d",&u,&v,&w);
g1[u][v]=w;
g2[v][u]=w;
}
dijkstra(g1,dis1);
dijkstra(g2,dis2);
int ans=0;
for(i=1;i<=n;i++)
{
int t=dis1[i]+dis2[i];
ans=ans>t?ans:t;
}
printf("%d\n",ans);
}
return 0;
}
poj 3268 Silver Cow Party （最短路）
原文：http://blog.csdn.net/u011721440/article/details/39010567