POJ - 3268 Silver Cow Party

最新推荐文章于 2023-07-07 15:49:51 发布

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本文解决了一个有趣的问题：如何计算一群牛从各自的农场出发参加聚会再返回所需的最长总时间。通过两次使用Dijkstra算法来确定每头牛到达聚会地点及返回的最短路径。

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
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: Ai, Bi, and Ti. The described road runs from farm Ai to farm Bi, requiring Ti 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.
给定一些边和距离且是单向的,然后N头牛,所有的按照最短路走到X处,并用最短路返回,问这些牛中所有路线里最大值是多少。写两个dijkstra，分别计算i到x的最短路，x到i的最短路。最后在枚举一下就可以了

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#define inf 1<<29

using namespace std;
int n,m,x;
int w[1010][1010];
int d1[1010],d2[1010];
int vis[1010];

void dijkstra1()//x到i的最短路
{
    memset(vis,0,sizeof(vis));
    for(int i=1;i<=n;i++)
    {
        d1[i]=w[x][i];
    }
    for(int k=1;k<=n;k++)
    {
        int a,min1=inf;
        for(int i=1;i<=n;i++)
        {
            if(!vis[i]&&min1>d1[i])
            {
                a=i;
                min1=d1[i];
            }
        }
        vis[a]=1;
        for(int i=1;i<=n;i++)
        {
            if(!vis[i]&&w[a][i]+d1[a]<d1[i])
            {
                d1[i]=d1[a]+w[a][i];
            }
        }
    }
}
void dijkstra2()//i到x的最短路
{
    memset(vis,0,sizeof(vis));
    for(int i=1;i<=n;i++)
    {
        d2[i]=w[i][x];
    }
    for(int k=1;k<=n;k++)
    {
        int a,min1=inf;
        for(int i=1;i<=n;i++)
        {
            if(!vis[i]&&min1>d2[i])
            {
                a=i;
                min1=d2[i];
            }
        }
        vis[a]=1;
        for(int i=1;i<=n;i++)
        {
            if(!vis[i]&&w[i][a]+d2[a]<d2[i])
            {
                d2[i]=d2[a]+w[i][a];
            }
        }
    }
}

int main()
{
    while(scanf("%d%d%d",&n,&m,&x)==3)
    {
        for(int i=1;i<=n;i++)
        {
            for(int j=1;j<=n;j++)
            {
                if(i==j)
                {
                    w[i][j]=0;
                }
                else
                {
                    w[i][j]=inf;
                }
            }
        }
        int a,b,c;
        for(int i=1;i<=m;i++)
        {
            scanf("%d%d%d",&a,&b,&c);
            w[a][b]=c;
        }
        dijkstra1();
        dijkstra2();
        int sum=-1;
        for(int i=1;i<=n;i++)
        {
            if(sum<d1[i]+d2[i])
            {
                sum=d1[i]+d2[i];
            }
        }
        printf("%d\n",sum);
    }
    return 0;
}