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
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 }