poj 3268 Silver Cow Party

最新推荐文章于 2021-04-01 18:48:34 发布

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本文深入探讨了深度学习在人工智能领域的应用，包括卷积神经网络、循环神经网络、强化学习等关键技术，以及它们在图像处理、自然语言处理、推荐系统等实际场景中的实践案例。

Silver Cow Party

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 10102		Accepted: 4489

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; roadi 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?

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 farmA_i to farmB_i, requiring T_i 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.

题意：一个有向图，求每个点：到终点的距离+终点到该点的距离，从中找出最大值。

思路：求终点到每个点的距离用一次dij就可以求出，但是求每个点到终点的距离时，如果进行n次dij，则会TLE。解决方法是先在原图上用一次dij求出终点到每个点的距离，然后在反图上再用一次dij求出终点到每个点的距离。这样只用了两次dij。

AC代码：

#include <iostream>
#include <cstring>
#include <string>
#include <cstdio>
#include <algorithm>
#include <queue>
#include <cmath>
#include <vector>
#include <cstdlib>

using namespace std;
const int INF=10000000;
int n,x;
int map[1005][1005];
int dis[1005];
bool vis[1005];
int back[1005];     //保存回去时终点到各点的最短距离
int go[1005];       //保存各点到终点的最短距离
void dij(int d[])
{
    memset(vis,false,sizeof(vis));
    for(int i=1;i<=n;i++)
    dis[i]=map[x][i];
    vis[x]=true;
    for(int i=1;i<=n;i++)
    {
        int x,min=INF;
        for(int j=1;j<=n;j++)
        if(!vis[j]&&dis[j]<min)
        min=dis[x=j];
        if(min==INF) break;
        vis[x]=true;
        for(int j=1;j<=n;j++)
        if(!vis[j]&&dis[j]>dis[x]+map[x][j])
            dis[j]=dis[x]+map[x][j];
    }
    for(int i=1;i<=n;i++)
    d[i]=dis[i];
}
int main()
{
   int m,a,b,t;
   while(scanf("%d%d%d",&n,&m,&x)!=EOF)
   {
       for(int i=1;i<=n;i++)
       for(int j=1;j<=n;j++)
       {
           if(i==j) map[i][j]=0;
           else
           map[i][j]=INF;
       }
       while(m--)
       {
           scanf("%d%d%d",&a,&b,&t);
           map[a][b]=t;
       }
       dij(back);
       for(int i=1;i<=n;i++)       //将矩阵转置
       for(int j=i+1;j<=n;j++)
       {
           int temp=map[i][j];
           map[i][j]=map[j][i];
           map[j][i]=temp;
       }
       dij(go);
       int max=0;
       for(int i=1;i<=n;i++)
       {
           if(back[i]+go[i]>max)
           max=back[i]+go[i];
       }
       printf("%d\n",max);
   }
   return 0;
}