題目背景

John的農場缺水了！！！

題目描述

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

農民John 決定將水引入到他的n(1<=n<=300)個牧場。他準備通過挖若

乾井，並在各塊田中修築水道來連通各塊田地以供水。在第i 號田中挖一口井需要花費W_i(1<=W_i<=100,000)元。連線i 號田與j 號田需要P_ij (1 <= P_ij <= 100,000 , P_ji=P_ij)元。

請求出農民John 需要為連通整個牧場的每一塊田地所需要的錢數。

輸入輸出格式

第1 行為一個整數n。

第2 到n+1 行每行一個整數，從上到下分別為W_1 到W_n。

第n+2 到2n+1 行為一個矩陣，表示需要的經費（P_ij）。

只有一行，為一個整數，表示所需要的錢數。

第一眼以為是個DP  但是發現有後效性

有點像網路流的思想   我們得有一個源泉 那就是井  而且必須有一口井

所以我們把井當做一個點  把它與所有點連邊  邊權為在那個點打井的費用

然後各個點連引水邊

跑一次最小生成樹 就OK了

原始碼

#include <bits/stdc++.h>
 

using namespace std;

int top, n, fa[305], fe[305], mp[305][305], head[305];

struct Node
{
    int x, y, nxt, w;
    Node() {    }
    Node( int x, int y, int w, int nxt ) : x(x), y(y), w(w), nxt(nxt) {    }
    inline bool operator < ( const Node &a ) const
    {
        return w < a.w;
    }
} e[150005];

inline void Adde( int x, int y, int w )
{
    e[++top] = Node(x, y, w, head[x]), head[x] = top;
    e[++top] = Node(y, x, w, head[y]), head[y] = top;
}

int Find( int x )
{
    return x == fa[x] ? x : fa[x] = Find(fa[x]);
}

int main()
{
    cin >> n;
    for(int i = 1; i <= n; ++i) 
        fa[i] = i, scanf( "%d", &fe[i] );
    for(int i = 1; i <= n; ++i)
        for(int j = 1; j <= n; ++j)
            scanf( "%d", &mp[i][j] );
    for(int i = 1; i <= n; ++i)
        for(int j = i + 1; j <= n; ++j)
            Adde(i, j, mp[i][j]);
    for(int i = 1; i <= n; ++i) Adde(0, i, fe[i]);
    sort(e + 1, e + top + 1);
    long long ans = 0;    
    for(int i = 1; i <= top; ++i)
    {
        int u = Find(e[i].x), v = Find(e[i].y);
        if(u != v)
        {
            fa[u] = v;
            ans += e[i].w;
        }
    }
    cout << ans << endl;
    return 0;
}