把联通每个农田最小的花费放在p[]里，要是联通的花费比挖井大就直接挖井，要是最后发现一口井都没挖，就找挖一口井的花费和联通它的花费差最小的，因为如果挖井价格比联通它小，那前面就会确定是挖井而不是联通了，所有这里差最小就代表多花钱最少的做法，#1#8过了，别的都wa

#include<iostream>
#include<algorithm>
#include<cstring>
using namespace std;
const int N = 3e5 + 10;
int p[N];
int nt[400][400];
int sj[400];
int bj[400];
int main() {
    int n;
    cin >> n;
    for (int i = 0; i < n; i++)
    {
        p[i] = N;
        cin >> sj[i];
    }
    for (int i = 0; i < n; i++)
    {
        for (int j = 0; j < n; j++)
        {
            cin >> nt[i][j];
        }
    }
    for (int i=0; i < n; i++)
    {
        int t = -1;
        for (int j = 0; j < n; j++)
        {
            if (nt[i][j]!=0&&!bj[j] && (t == -1 || p[t] > p[j]))
            {
                t = j;
            }
        }
        if (t == -1)continue;
        bj[t] = 1;
        for (int j = 0; j < n; j++)
        {
            if(nt[t][j]!=0)p[j] = min(p[j], nt[t][j]);
        }
    }
    int minn=N;
    int flag = 0;
    for (int i = 0; i < n; i++)
    {
        minn = min(minn, abs(p[i] - sj[i]));
        p[i] = min(p[i], sj[i]);
        if (p[i] == sj[i])flag = 1;
    }
    int ans = 0;
    for (int i = 0; i < n; i++)
    {
        ans += p[i];
    }
    if (flag)cout << ans << endl;
    else cout << ans + minn;
    return 0;
}