Silver Cow Party

最新推荐文章于 2019-07-30 21:30:52 发布

原创最新推荐文章于 2019-07-30 21:30:52 发布 · 114 阅读

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本文介绍了一个关于寻找最优路径的问题，具体为多个农场中每只奶牛到特定农场参加聚会并返回的最短时间问题。通过两次Dijkstra算法计算往返所需的最大时间。

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Silver Cow Party

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 requiresT_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?

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.

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

Sample Input

Sample Output

题目大意：有n个农场，从每个农场出来一只奶牛到第x个农场参加聚会，参加完聚会后返回，每条路是单向的，所以不能原路返回，问每头奶牛需要的时间中的最长时间；

思路：起点不一样，终点一样，所以从终点开始找到每个农场的最短路，因为是从起点到终点，所有要把路径反向；两次dijkstra后，算出来回需要的最大值；

#include<stdio.h>
#include<string.h>
#include<algorithm>
#define inf 0x3f3f3f3f
#define ll long long
using namespace std;
int e[1005][1005];//用来储存正向路径
int ee[1005][1005];//用来储存反向路径
int mark[1005];
int dis[1005],ds[1005];

int main()
{
    int n,m,x;
    while(~scanf("%d %d %d",&n,&m,&x))//n个农场，m条路，到x农场开聚会
    {
        memset(e,0,sizeof(e));
        memset(ee,0,sizeof(ee));
        memset(dis,0,sizeof(dis));
        memset(ds,0,sizeof(ds));
        memset(mark,0,sizeof(mark));
        int a,b,l,i,j;
        for(i=1; i<=n; i++)
            for(j=1; j<=n; j++)
                if(i==j)
                {
                    e[i][j]=0;
                    ee[i][j]=0;
                }
                else
                {
                    e[i][j]=inf;
                    ee[i][j]=inf;
                }
        for(i=1; i<=m; i++)
        {
            scanf("%d %d %d",&a,&b,&l);
            e[a][b]=min(e[a][b],l);
            ee[b][a]=min(ee[b][a],l);
        }
        int pos=x,s=0;
        for(i=1;i<=n;i++)
            ds[i]=ee[pos][i];
        mark[pos]=1;
        for(i=1; i<n; i++)
        {
            int mini=inf;
            for(j=1; j<=n; j++)
            {
                if(mark[j]==0&&ds[j]<mini)
                {
                    pos=j;
                    mini=ds[j];
                }
            }
            mark[pos]=1;
            for(j=1; j<=n; j++)
            {
                if(mark[j]==0&&ds[j]>ds[pos]+ee[pos][j])
                    ds[j]=ee[pos][j]+ds[pos];
            }
        }
        pos=x;
        memset(mark,0,sizeof(mark));
        mark[pos]=1;
        for(i=1; i<=n; i++)
            dis[i]=e[pos][i];
        for(i=1; i<n; i++)
        {
            int mini=inf;
            for(j=1; j<=n; j++)
            {
                if(mark[j]==0&&dis[j]<mini)
                {
                    pos=j;
                    mini=dis[j];
                }
            }
            mark[pos]=1;
            for(j=1; j<=n; j++)
            {
                if(mark[j]==0&&dis[j]>e[pos][j]+dis[pos])
                    dis[j]=e[pos][j]+dis[pos];
            }
        }
        for(i=1;i<=n;i++)
        {
            if(ds[i]+dis[i]>s)
                s=dis[i]+ds[i];
        }
        printf("%d\n",s);
    }
}