POJ - 3268 Silver Cow Party【dijstra】

最新推荐文章于 2021-05-04 13:37:00 发布

原创最新推荐文章于 2021-05-04 13:37:00 发布 · 133 阅读

0 ·

CC 4.0 BY-SA版权

图论专栏收录该内容

24 篇文章

订阅专栏

每个编号为1到N的农场中有一头奶牛将参加在农场X（1≤X≤N）举行的大型奶牛聚会。存在M条单向道路连接这些农场，每条道路i需要Ti（1≤Ti≤100）单位时间通过。奶牛会选择时间最短的路线，返回路线可能与前往路线不同。找出所有奶牛中花费最长步行时间去聚会并返回的奶牛的时间是多少。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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

Sample Output

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.

#include<iostream>
#include<cstring>
#include<algorithm>
using namespace std;
const int INF = 0x3f3f3f3f;
int N,M,X;
int Map[1005][1005];
bool vis[1005];
int dis[1005];
int sum[1005];
void Reverse()
{
    for (int i =1; i <= N; i++)
    {
        for (int j = i; j <= N; j++)
        {
            swap(Map[i][j], Map[j][i]);
        }
    }
}
void dijstra(int s)
{
    memset(vis, false, sizeof(vis));
    for (int i = 1; i <= N; i++)
        dis[i] = Map[s][i];
    dis[s] = 0;
    vis[s] = true;
    for (int i = 1; i < N; i++)
    {
        int Min = INF, k = 0;
        for (int j = 1; j <= N; j++)
        {
            if(!vis[j] && Min > dis[j])
            {
                Min = dis[j];
                k = j;
            }
        }
        if(k == 0) break;
        vis[k] = true;
        for (int j = 1; j <= N; j++)
        {
            if(!vis[j] && dis[j] > Map[k][j] + dis[k])
                dis[j] = Map[k][j] + dis[k];
        }
    }
   for (int i = 1; i <= N; i++)
   {
       sum[i] += dis[i];
   }
}
int main()
{
    while(scanf("%d%d%d", &N, &M, &X) != EOF)
    {
        for (int i = 1; i <= N; i++)
          for(int j = 1; j <= N; j++)
          {
            Map[i][j] = INF;

          }
        while(M--)
        {
            int u, v, w;
            cin >> u >> v >> w;
            Map[u][v] = w;
        }
        dijstra(X);
        Reverse();
        dijstra(X);
        sort(sum + 1, sum + N + 1);
        cout << sum[N] << endl;
    }

}