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
1 #include<iostream>
2 #include<cstring>
3 #include<cstdio>
4 using namespace std;
5 int f[310][310],dis[310];
6 bool vis[310];
7 int main()
8 {
9 memset(f,0x3f,sizeof(f));
10 memset(dis,0x3f,sizeof(dis));
11 int n;
12 scanf("%d",&n);
13 for(int i=1;i<=n;++i)
14 scanf("%d",&dis[i]);
15 for(int i=1;i<=n;++i)
16 for(int j=1;j<=n;++j)
17 scanf("%d",&f[i][j]);
18 for(int i=1;i<=n;++i)
19 {
20 int k=0;
21 for(int j=1;j<=n;++j)
22 if(!vis[j]&&dis[j]<dis[k])
23 k=j;
24 vis[k]=1;
25 for(int j=1;j<=n;++j)
26 if(j!=k&&!vis[j]&&dis[j]>f[k][j])
27 dis[j]=f[k][j];
28 }
29 int ans=0;
30 for(int i=1;i<=n;++i)
31 ans+=dis[i];
32 printf("%d",ans);
33 return 0;
34 }
原文地址：https://www.cnblogs.com/axma/p/9356912.html