Watering Hole
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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.
翻译
农民John 决定将水引入到他的边长为n(1<=n<=300)的牧场。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:
454430 2 2 22 0 3 32 3 0 42 3 4 0
输出样例#1:
9
说明
John等着用水,你只有1s时间!!!
用最少的花费把所有点都连起来,最小生成树……井的话就看做一个超级源点0号点,这就成了求 0 - n 这 n + 1 个点的最小生成树。
代码如下
#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<queue>using namespace std;const int size = 200010;int n;int num[size];struct dc{ int f,t,d;}l[size];int tot = 1;void build(int f,int t,int d){ l[tot].f = f; l[tot].t = t; l[tot].d = d; tot ++;}int f[size];int find(int x){ if(f[x] == x) return x; return f[x] = find(f[x]);}bool cmp(dc a,dc b){ return a.d < b.d;}int main(){ scanf("%d",&n); for(int i = 1 ; i <= n ; i ++) scanf("%d",&num[i]) , f[i] = i; for(int i = 1 ; i <= n ; i ++) for(int j = 1 ; j <= n ; j ++) { int d; scanf("%d",&d); build(i,j,d); } for(int i = 1 ; i <= n ; i ++) build(0,i,num[i]); sort(l+1,l+tot,cmp); int ans = 0; for(int i = 1 ; i < tot ; i ++) { int ff = l[i].f , ft = l[i].t; if(find(ff) != find(ft)) { f[find(ff)] = find(ft); ans += l[i].d; } } printf("%d\n",ans); return 0;}
传送门:https://www.luogu.org/problem/show?pid=1550
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