luogu P1550 [USACO08OCT]打井Watering Hole

题目背景

John的农场缺水了！！！

题目描述

Farmer John has decided to bring water to his N (1 <= N <= 300) pastures which are conveniently numbered 1..N. He may bring water to a pasture either by building a well in that pasture or connecting the pasture via a pipe to another pasture which already has water.

Digging a well in pasture i costs W_i (1 <= W_i <= 100,000).

Connecting pastures i and j with a pipe costs P_ij (1 <= P_ij <= 100,000; P_ij = P_ji; P_ii=0).

Determine the minimum amount Farmer John will have to pay to water all of his pastures.

POINTS: 400

农民John 决定将水引入到他的n(1<=n<=300)个牧场。他准备通过挖若

干井，并在各块田中修筑水道来连通各块田地以供水。在第i 号田中挖一口井需要花费W_i(1<=W_i<=100,000)元。连接i 号田与j 号田需要P_ij (1 <= P_ij <= 100,000 , P_ji=P_ij)元。

请求出农民John 需要为连通整个牧场的每一块田地所需要的钱数。

输入输出格式

输入格式：

第1 行为一个整数n。

第2 到n+1 行每行一个整数，从上到下分别为W_1 到W_n。

第n+2 到2n+1 行为一个矩阵，表示需要的经费（P_ij）。

输出格式：

只有一行，为一个整数，表示所需要的钱数。

输入输出样例

输入样例#1：

4
5
4
4
3
0 2 2 2
2 0 3 3
2 3 0 4
2 3 4 0

输出样例#1：

9

说明

John等着用水，你只有1s时间！！！

开始给dis赋值打井花费来比较打井和修水道哪种更优

 #include<iostream>
 #include<cstring>
 #include<cstdio>
 using namespace std;
 int f[][],dis[];
 bool vis[];
 int main()
 {
     memset(f,0x3f,sizeof(f));
     memset(dis,0x3f,sizeof(dis));
     int n;
     scanf("%d",&n);
     for(int i=;i<=n;++i)
         scanf("%d",&dis[i]);
     for(int i=;i<=n;++i)
         for(int j=;j<=n;++j)
             scanf("%d",&f[i][j]);
     for(int i=;i<=n;++i)
     {
         int k=;
         for(int j=;j<=n;++j)
             if(!vis[j]&&dis[j]<dis[k])
                 k=j;
         vis[k]=;
         for(int j=;j<=n;++j)
             if(j!=k&&!vis[j]&&dis[j]>f[k][j])
                 dis[j]=f[k][j];
     }
     int ans=;
     for(int i=;i<=n;++i)
         ans+=dis[i];
     printf("%d",ans);
     return ;
 }