[洛谷P1550] [USACO08OCT]打井Watering Hole [最小生成树]

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。

第一眼以为是个DP  但是发现有后效性

有点像网络流的思想   我们得有一个源泉 那就是井  而且必须有一口井

所以我们把井当做一个点  把它与所有点连边  边权为在那个点打井的费用

然后各个点连引水边

跑一次最小生成树 就OK了

源代码

#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;}