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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
题解：还是正向建图和逆向建图。就是求往返路程中最大的一条。
djistra：
#include <iostream>
#include <cstring>
#include <cstdio>
#include <queue>
using namespace std;
const int INF= 0x3fffffff;
int map[2][1003][1003];
bool visited[1003];
int d[1003];
int ans[1003];
int max(int a,int b)
{
return a > b ? a : b;
}
void prim(int n,int s,int flag)
{
memset(visited,false,sizeof(visited));
for(int i = 1;i <= n;i++)
{
d[i] = map[flag][s][i];
//cout<<d[i]<<endl;
}
visited[s] = true;
for(int i = 1;i < n;i++)
{
int min = INF;
int k;
for(int j = 1;j <= n;j++)
{
if(!visited[j] && min > d[j])
{
min = d[j];
k = j;
}
}
if(min == INF)
{
break;
}
visited[k] = true;
for(int j = 1;j <= n;j++)
{
if(!visited[j] && d[j] > d[k] + map[flag][k][j])
{
d[j] = d[k] + map[flag][k][j];
}
}
}
}
int main()
{
int n,m,st;
while(scanf("%d%d%d",&n,&m,&st) != EOF)
{
for(int i = 1;i <= n;i++)
{
for(int j = 1;j <= n;j++)
{
if(i == j)
{
map[0][i][j] = 0;
map[1][i][j] = 0;
}
else
{
map[0][i][j] = INF;
map[1][i][j] = INF;
}
}
}
int u,v,c;
for(int i = 0;i < m;i++)
{
scanf("%d%d%d",&u,&v,&c);
map[0][u][v] = c;
map[1][v][u] = c;
}
prim(n,st,1);
for(int i = 1;i <= n;i++)
{
ans[i] = d[i];
}
prim(n,st,0);
int res = 0;
for(int i = 1;i <= n;i++)
{
ans[i] += d[i];
res = max(res,ans[i]);
}
printf("%d\n",res);
}
return 0;
}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/wang2534499/article/details/47753441
