bzoj 3479: [Usaco2014 Mar]Watering the Fields

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3479: [Usaco2014 Mar]Watering the FieldsTime Limit: 10 Sec  Memory Limit: 128 MB
Submit: 174  Solved: 97
[Submit][Status][Discuss]Description

 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.

草坪上有N個水龍頭，位於（xi,yi）

求將n個水龍頭連通的最小費用。
任意兩個水龍頭可以修剪水管，費用為歐幾裡得距離的平方。修水管的人只願意修費用大於等於c的水管。

 

Input

* Line 1: The integers N and C.

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

Output

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

Sample Input3 11
0 2
5 0
4 3

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.
Sample Output 46
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. HINT Source

Silver 譯文By Hta

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　　居然當成162M，結果數組開爆了。。。。。

 1 #include<bits/stdc++.h> 2 using namespace std; 3 const int MAX=2001; 4 struct node{ 5     int x,y; 6 }; 7 node pos[MAX]; 8 struct node1{ 9     int left,right;10     int cost;11 };12 node1 edge[MAX*MAX];13 int cmp(const node1 &a,const node1 &b){14     if(a.cost<b.cost) return 1;15     return 0;16 }17 int fa[MAX];18 int get_fa(int v){19     if(v!=fa[v]){20         fa[v]=get_fa(fa[v]);21     }22     return fa[v];23 }24 int hehe(int m,int n){25     if(m>n){26         m=m+n;27         n=m-n;28         m=m-n;29     }30     int mm=get_fa(m);31     int nn=get_fa(n);32     fa[mm]=nn; 33 }34 int N;35 int C;36 int totedge;37 int sum;38 int tot;39 int main(){40     41     cin>>N>>C;42     for(int i=1;i<=N;i++){43         int a,b;44         cin>>a>>b;45         pos[i].x=a;46         pos[i].y=b;47     }48     49     for(int i=1;i<=N;i++){50         for(int j=1;j<=N;j++){51             edge[++totedge].left=i;52             edge[totedge].right=j;53             edge[totedge].cost=(pos[i].x-pos[j].x)*(pos[i].x-pos[j].x)+54                             (pos[i].y-pos[j].y)*(pos[i].y-pos[j].y);55         }56     }57     58     for(int i=1;i<=N;i++) fa[i]=i;59     sort(edge+1,edge+totedge+1,cmp);60     for(int i=1;i<=totedge;i++){61         if(edge[i].cost>=C){62             int h=edge[i].left;63             int g=edge[i].right;64             if(get_fa(h)!=get_fa(g)){65                 hehe(h,g);66                 sum+=edge[i].cost;67                 tot++;68                 if(tot==N-1) break;69             }70         }71     }72     73     if(tot==N-1) cout<<sum;74     if(tot<N-1) cout<<-1;75     return 0;76 } 

 

bzoj 3479: [Usaco2014 Mar]Watering the Fields