| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 15460 | Accepted: 7004 |
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
题目大意：
n头牛要到达某一个点X，然后再返回到原来的位置。同时要求对于每头牛路径必须是最短的，而且去的路和回来的路不一样。求这些牛所花费的最长时间。参考代码:
#include<stack>
#include<queue>
#include<cmath>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
const double eps=1e-10;
const int INF=0x3f3f3f3f;
const int MAXN=1100;
int n,m,p,edge[MAXN][MAXN],backedge[MAXN][MAXN],mincost1[MAXN],mincost2[MAXN];
bool used1[MAXN],used2[MAXN];
void dijkstra(int start)
{
for(int i=1; i<=n; i++)
{
used1[i]=used2[i]=false;
mincost1[i]=edge[start][i];
mincost2[i]=edge[i][start];
}
used1[start]=used2[start]=true;
int v;
while(true)
{
v=-1;
for(int i=1; i<=n; i++)
if(!used1[i]&&(v==-1||mincost1[i]<mincost1[v]))
v=i;
if(v==-1)
break;
used1[v]=true;
for(int i=1; i<=n; i++)
if(!used1[i]&&mincost1[i]>mincost1[v]+edge[v][i])
mincost1[i]=mincost1[v]+edge[v][i];
v=-1;
for(int i=1; i<=n; i++)
if(!used2[i]&&(v==-1||mincost2[i]<mincost2[v]))
v=i;
if(v==-1)
break;
used2[v]=true;
for(int i=1; i<=n; i++)
if(!used2[i]&&mincost2[i]>mincost2[v]+backedge[v][i])
mincost2[i]=mincost2[v]+backedge[v][i];
}
}
int main()
{
#ifndef ONLINE_JUDGE
freopen("in.txt","r",stdin);
#endif // ONLINE_JUDGE
while(scanf("%d%d%d",&n,&m,&p)!=EOF)
{
for(int i=1; i<=n; i++)
for(int j=1; j<=n; j++)
if(i==j)
edge[i][j]=backedge[i][j]=0;//方向相反
else
edge[i][j]=backedge[i][j]=INF;
for(int i=1; i<=m; i++)
{
int a,b,c;
scanf("%d%d%d",&a,&b,&c);
edge[b][a]=backedge[a][b]=c;
}
dijkstra(p);
int ans=-1;
for(int i=1; i<=n; i++)
ans=max(ans,mincost1[i]+mincost2[i]);
printf("%d\n",ans);
}
return 0;
}
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POJ 3268 Silver Cow Party（dijkstra+矩阵转置）
原文地址：http://blog.csdn.net/noooooorth/article/details/47702069
