Silver Cow Party---poj3268

Silver Cow Party

Time Limit:2000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

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 T_i (1 ≤ T_i ≤ 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

Line 1: Three space-separated integers, respectively: N, M, and X
Lines 2.. M+1: Line i+1 describes road i with three space-separated integers: A_i, B_i, and T_i. The described road runs from farm A_i to farm B_i, requiring T_itime units to traverse.

Output

Line 1: One integer: the maximum of time any one cow must walk.

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

Cow 4 proceeds directly to the party (3 units) and returns via farms 1 and 3 (7 units), for a total of 10 time units.

题意: 有n个农场，将在农场x举行party，农场a->b用时c；有向图；求所有农场中的牛去到农场x再回去所需的最长时间；可以两次运用Dij算法，用dist[i]表示起点到i

的最短时间

#include<iostream>
#include<stdio.h>
#include<math.h>
#include<string.h>
#include<algorithm>
#define INF 0xfffffff
#define N 1100
using namespace std;
#define max(a,b) (a>b?a:b)
#define min(a,b) (a<b?a:b)
int x,n,m,dist1[N],dist2[N],vis[N];
int maps[N][N];

void Dij1()
{
    int i,j;

    vis[x]=0;

    dist1[x]=0;

    for(i=1; i <= n; i++)
    {
        int Min=INF,index=-1;

        for(j = 1; j <= n; j++)
        {
            if(vis[j] == 0 && Min > dist1[j])
            {
                Min = dist1[j];
                index = j;
            }
        }
        if(index==-1)break;
        vis[index]=1;

        for(j = 1; j <= n; j++)
            dist1[j] = min(dist1[j],dist1[index] + maps[index][j]);
    }
}
void Dij2()
{
    int i,j;

    vis[x]=0;

    dist2[x]=0;

    for(i=1; i <= n; i++)
    {
        int Min=INF,index=-1;

        for(j = 1; j <= n; j++)
        {
            if(vis[j] == 0 && Min > dist2[j])
            {
                Min = dist2[j];
                index = j;
            }
        }
        if(index==-1)break;
        vis[index]=1;

        for(j = 1; j <= n; j++)
            dist2[j] = min(dist2[j],dist2[index] + maps[j][index]);
    }
}

int main()
{
    int i,a,b,c;
    while(scanf("%d%d%d",&n,&m,&x)!=EOF)
    {
        for(i=0;i<=n;i++)
        {
            dist1[i]=dist2[i]=INF;
            for(int j=1;j<=n;j++)
                maps[i][j]=INF;
        }
        while(m--)
        {
            scanf("%d%d%d", &a, &b, &c);

            maps[a][b] = c;
        }

        memset(vis,0,sizeof(vis));
        Dij1();//去

        memset(vis,0,sizeof(vis));
        Dij2();//回来；

        int ans=0;

        for(i=1;i<=n;i++)
        {
            ans=max(ans,dist1[i]+dist2[i]);
        }
        printf("%d\n",ans);
    }
    return 0;
}