POJ 3268 Silver Cow Party（兩次Dijkstra求最大值）

Silver Cow Party

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 13489		Accepted: 6075

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_i time 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.

Source

題目大意：

說的是有N個農場，每個農場出一個奶牛去第x個農場參加派對，i個弄農場到i+1個農場有一個距離，先求出每個奶牛從自己的農場出來後到達x個農場參加party再返回自己所在的農場的最短路，然後在這些最短的裏面找到最大的一條。

解題思路：

其實就是Dijkstra的簡單應用不過要注意的是，數據太大，用O（n3）的Floyd會T，所以呢，只要用兩次Dijkstra就可以了，第一次以x爲源點，求出其餘的n-1個農場到達x的最短路，然後，再將edge[i][j]的行和列相互互換位置，就得到了n-1個農場到第x個農場的距離，然後再用一次Dijkstra，求出了第n-1個農場到第x個農場的距離，然後求他們的和的最大值就OK了。

代碼：

# include<cstdio>
# include<iostream>
# include<cstring>

using namespace std;

# define MAX 1000+4
# define inf 99999999

int edge[MAX][MAX];
int book[MAX];
int dis[MAX];
int bdis[MAX];
int n,m,x;
int u,v;


void init()
{
    for ( int i = 1;i <= n;i++ )
    {
        for ( int j = 1;j <= n;j++ )
        {
            if ( i==j )
            {
                edge[i][j] = 0;
            }
            else
            {
                edge[i][j] = inf;
            }
        }
    }
}

void input()
{
    for ( int i = 1;i <= m;i++ )
    {
        int t1,t2,t3;
        cin>>t1>>t2>>t3;
        edge[t1][t2] = t3;
    }
}

int Dijkstra()
{
    for ( int i = 1;i <= n;i++ )
    {
        book[i] = 0;
        dis[i] = edge[x][i];
        bdis[i] = edge[i][x];
    }

    int _min;
    book[x] = 1;

    for ( int i = 1;i <= n-1;i++ )
    {
         _min = inf;
         for ( int j = 1;j <= n;j++ )
         {
             if ( book[j]==0&&dis[j] < _min )
             {
                 _min = dis[j];
                 u = j;
             }
         }


    book[u] = 1;
    for ( v = 1;v <= n;v++ )
    {
        if ( book[v]==0&&dis[v] > dis[u]+edge[u][v] )
        {
            dis[v] = dis[u]+edge[u][v];
        }
    }

    }

    memset(book,0,sizeof(book));

    book[x] = 1;
    for ( int i = 1;i <= n;i++ )
    {
        _min = inf;
        for ( int j = 1;j <= n;j++ )
        {
            if ( book[j]==0 && bdis[j] < _min )
            {
                _min = bdis[j];
                 u = j;
            }
        }

        book[u] = 1;
        for ( v = 1;v <= n;v++ )
        {
            if ( book[v]==0&&bdis[v] > bdis[u]+edge[v][u] )
            {
                bdis[v] = bdis[u]+edge[v][u];
            }
        }
    }
        _min = -1;
        for ( int i = 1;i <= n;i++ )
        {
            if ( bdis[i]+dis[i] > _min )
            {
                _min = bdis[i]+dis[i];
            }
        }

    return _min;

}

int main(void)
{
        cin>>n>>m>>x;
        init();
        input();
        printf("%d\n",Dijkstra());
        return 0;
}