【洛谷 2212】浇地Watering the Fields

太过爱你忘了你带给我的痛 2021-10-24 04:54 96阅读 0赞

### 题目描述 ###

Due to a lack of rain, Farmer John wants to build an irrigation system to

send 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 two

fields 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 his

fields are linked together -- so that water in any field can follow a

sequence of pipes to reach any other field.

Unfortunately, the contractor who is helping FJ install his irrigation

system refuses to install any pipe unless its cost (squared Euclidean

length) 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NN (1 <= NN <= 2000)块田送水。农田在一个二维平面上，第i块农田坐标为(x\_ixi , y\_iyi )（0 <= x\_ixi , y\_iyi <= 1000），在农田ii 和农田jj 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(xi−xj)2\+(yi−yj)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 <= CC <= 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 \#1复制

3 11
    0 2
    5 0
    4 3

输出 \#1复制

### 说明/提示 ###

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

题解：仍然是板子kruskal，不过我jjj还是漏洞百出嘤嘤嘤

90分疯狂找错，最后发现没输出-1……我佛了

#include<iostream>
    #include<algorithm>
    #include<queue>
    #include<cmath>
    #include<cstring>
    #include<cstdlib>
    #include<cstdio>
    using namespace std;
    typedef double db;
    const int N=2005;
    const int MX=4000006;
    int n,m,cp,tot,fa[N],b[N];
    
    struct node{
        int x,y;
    }a[N];
    
    struct YCLL{
        int u,v;
        int va;
    }e[MX];
    
    int ans=0;
    
    int how_long(int uu,int vv){
        int x5=a[uu].x; int x6=a[vv].x;
        int y5=a[uu].y; int y6=a[vv].y;
        return (x5-x6)*(x5-x6)+(y5-y6)*(y5-y6);
    }
    
    bool cmp(YCLL aa,YCLL bb){
        return aa.va<bb.va;
    }
    
    int find(int x){
        if(x!=fa[x]) 
           fa[x]=find(fa[x]);
        return fa[x];
    }
    //int b[N];
    int main(){
        freopen("2212.in","r",stdin);
        freopen("2212.out","w",stdout);
        scanf("%d",&n); scanf("%d",&m);
        //for(int i=1;i<=n;i++) scanf("%d",&b[i]);
        for(int i=1;i<=n;i++) fa[i]=i;
        
        for(int i=1;i<=n;i++)
            scanf("%d %d",&a[i].x,&a[i].y);
        for(int i=1;i<=n;i++){
            for(int j=i+1;j<=n;j++){
                int oh=how_long(i,j);
                if(oh<m) continue;
                cp++; e[cp].u=i; e[cp].v=j;
                e[cp].va=oh;
                //cout<<e[cp].va<<endl;
            }
        }
        sort(e+1,e+cp+1,cmp);
        //cout<<cp;
        for(int i=1;i<=cp;i++){
            //cout<<"00340204"<<endl;
            int uu=find(e[i].u);
            int vv=find(e[i].v);
            if(uu==vv) continue;
            ans+=e[i].va; 
            fa[uu]=vv; tot++;
            if(tot==(n-1)) break;
        }
        if(tot==(n-1)) printf("%d",ans);
        else cout<<-1;
        return 0;
    }

转载于:https://www.cnblogs.com/wuhu-JJJ/p/11291333.html