[USACO14MAR]浇地Watering the Fields

题目描述

Due to a lack of rain, Farmer John wants to build an irrigation system tosend water between his N fields (1 <= N <= 2000).Each field i is described by a distinct point (xi, yi) in the 2D plane,

with 0 <= xi, yi <= 1000. The cost of building a water pipe between twofields i and j is equal to the squared Euclidean distance between them:(xi - xj)^2 + (yi - yj)^2

FJ would like to build a minimum-cost system of pipes so that all of hisfields are linked together -- so that water in any field can follow asequence of pipes to reach any other field.

Unfortunately, the contractor who is helping FJ install his irrigationsystem refuses to install any pipe unless its cost (squared Euclideanlength) is at least C (1 <= C <= 1,000,000).

Please help FJ compute the minimum amount he will need pay to connect all his fields with a network of pipes.

农民约翰想建立一个灌溉系统，给他的N (1 <= N <= 2000)块田送水。农田在一个二维平面上，第i块农田坐标为(x_i , y_i)（0 <= x_i , y_i <= 1000），在农田i 和农田j自己铺设水管的费用是这两块农田的欧几里得距离的平方(x_i - x_j)^2 + (y_i - y_j)^2。

农民约翰希望所有的农田之间都能通水，而且希望花费最少的钱。但是安装工人拒绝安装费用小于C的水管(1 <= C<= 1,000,000)。

请帮助农民约翰建立一个花费最小的灌溉网络，如果无法建立请输出-1。

输入输出格式

输入格式：

* Line 1: The integers N and C.

* Lines 2..1+N: Line i+1 contains the integers xi and yi.

输出格式：

* Line 1: The minimum cost of a network of pipes connecting the fields, or -1 if no such network can be built.

输入输出样例

输入样例#1：

3 11
0 2
5 0
4 3

输出样例#1：

46

说明

INPUT DETAILS:

There are 3 fields, at locations (0,2), (5,0), and (4,3). The contractor will only install pipes of cost at least 11.

OUTPUT DETAILS:

FJ cannot build a pipe between the fields at (4,3) and (5,0), since its cost would be only 10. He therefore builds a pipe between (0,2) and (5,0) at cost 29, and a pipe between (0,2) and (4,3) at cost 17.

Source: USACO 2014 March Contest, Silver

分析：

本题就是一道裸的最小生成树，只需要计算一下每两个点间的费用，并判断与c的关系，进而决定是否加边，但是注意数组是2000*2000，而不是2000，否则会RE。

CODE：

 #include<iostream>
 #include<cstdio>
 #include<algorithm>
 using namespace std;
 const int M=;
 int n,c,tot,ans;
 int fa[M];
 int xcor[M],ycor[M];
 struct node{
     int u,v,w;
 }a[M];
 int findr(int x){
     if (fa[x]==x) return x;
     return fa[x]=findr(fa[x]);
 }
 void merge(int x,int y){
     int A=findr(x);
     int B=findr(y);
     if (fa[A]!=fa[B]) fa[B]=A;
     return ;
 }
 bool cmp(node x,node y){return x.w<y.w;}
 int main(){
     cin>>n>>c;
     for (int i=;i<=n;i++) fa[i]=i;
     for(int i=;i<=n;i++) cin>>xcor[i]>>ycor[i];
     for (int i=;i<=n;i++){
         for (int j=i+;j<=n;j++){
             int cost=(xcor[i]-xcor[j])*(xcor[i]-xcor[j])+(ycor[i]-ycor[j])*(ycor[i]-ycor[j]);
             if (cost>=c){
                 a[++tot].u=i;
                 a[tot].v=j;
                 a[tot].w=cost;
             }
         }
     }
     sort(a+,a+tot+,cmp);
     int cnt=;
     for (int i=;i<=tot;i++){
         if (fa[findr(a[i].u)]!=fa[findr(a[i].v)]){
             ans+=a[i].w;
             merge(a[i].u,a[i].v);
             cnt++;
         }
     }
     if (cnt==n-) cout<<ans<<endl;
     else cout<<-<<endl;
     return ;
 }