Highways

Description

The island nation of Flatopia is perfectly flat. Unfortunately, Flatopia has a very poor system of public highways. The Flatopian government is aware of this problem and has already constructed a number of highways connecting some of the most important towns. However, there are still some towns that you can't reach via a highway. It is necessary to build more highways so that it will be possible to drive between any pair of towns without leaving the highway system.

Flatopian towns are numbered from 1 to N and town i has a position given by the Cartesian coordinates (xi, yi). Each highway connects exaclty two towns. All highways (both the original ones and the ones that are to be built) follow straight lines, and thus their length is equal to Cartesian distance between towns. All highways can be used in both directions. Highways can freely cross each other, but a driver can only switch between highways at a town that is located at the end of both highways.

The Flatopian government wants to minimize the cost of building new highways. However, they want to guarantee that every town is highway-reachable from every other town. Since Flatopia is so flat, the cost of a highway is always proportional to its length. Thus, the least expensive highway system will be the one that minimizes the total highways length.

Input

The input consists of two parts. The first part describes all towns in the country, and the second part describes all of the highways that have already been built.

The first line of the input file contains a single integer N (1 <= N <= 750), representing the number of towns. The next N lines each contain two integers, xi and yi separated by a space. These values give the coordinates of ith town (for i from 1 to N). Coordinates will have an absolute value no greater than 10000. Every town has a unique location.

The next line contains a single integer M (0 <= M <= 1000), representing the number of existing highways. The next M lines each contain a pair of integers separated by a space. These two integers give a pair of town numbers which are already connected by a highway. Each pair of towns is connected by at most one highway.

Output

Write to the output a single line for each new highway that should be built in order to connect all towns with minimal possible total length of new highways. Each highway should be presented by printing town numbers that this highway connects, separated by a space.

If no new highways need to be built (all towns are already connected), then the output file should be created but it should be empty.

Sample Input

9

1 5

0 0

3 2

4 5

5 1

0 4

5 2

1 2

5 3

3

1 3

9 7

1 2

Sample Output

1 6

3 7

4 9

5 7

8 3

题意

给你n个点的坐标，然后给你m条已知的边，求最小生成树

题解

先把所有的点存起来

然后把所有点互相连接的边枚举出来（所以MAXN得很大）

将已知的边先进行一次合并

然后常规Kruskal一下

由于不需要求权值和，只需要在每一次加边时候输出一下起点和终点

代码

#include<iostream>

#include<cstdio>     //EOF,NULL

#include<cstring>    //memset

#include<cstdlib>    //rand,srand,system,itoa(int),atoi(char[]),atof(),malloc

#include<cmath>           //ceil,floor,exp,log(e),log10(10),hypot(sqrt(x^2+y^2)),cbrt(sqrt(x^2+y^2+z^2))

#include<algorithm>  //fill,reverse,next_permutation,__gcd,

#include<string>

#include<vector>

#include<queue>

#include<stack>

#include<utility>

#include<iterator>

#include<iomanip>             //setw(set_min_width),setfill(char),setprecision(n),fixed,

#include<functional>

#include<map>

#include<set>

#include<limits.h>     //INT_MAX

#include<bitset> // bitset<?> n

using namespace std;

typedef long long ll;

typedef pair<int,int> P;

#define all(x) x.begin(),x.end()

#define readc(x) scanf("%c",&x)

#define read(x) scanf("%d",&x)

#define read2(x,y) scanf("%d%d",&x,&y)

#define read3(x,y,z) scanf("%d%d%d",&x,&y,&z)

#define print(x) printf("%d\n",x)

#define mst(a,b) memset(a,b,sizeof(a))

#define lowbit(x) x&-x

#define lson(x) x<<1

#define rson(x) x<<1|1

#define pb push_back

#define mp make_pair

const int INF =0x3f3f3f3f;

const int inf =0x3f3f3f3f;

const int mod = 1e9+;

const int MAXN =  + ;

const int maxn = ;

struct node{

  int st,ed,v;

  bool operator < (node b) const{

     return v < b.v;

  }

}rod[MAXN];

struct poot{

   int  x, y;

}pot[];

int n,m,v;

int cnt,ans;

int pre[MAXN];

int find(int x){ return x == pre[x] ? x : pre[x] = find(pre[x]);}

bool join(int x,int y){

    if(find(x)!=find(y)){

      pre[find(y)] = find(x);

      return true;

    }

    return false;

}

void kruskal(){

  for(int i =  ;i < cnt ; i++){

    int mp1 = find(rod[i].st);

    int mp2 = find(rod[i].ed);

    if(join(mp1,mp2)) printf("%d %d\n",rod[i].st, rod[i].ed);

  }

}

int main(){

        read(n);

        int a,b;

        for(int i = ; i <= n ;i++){

          read2(a,b);

          pot[i].x = a;

          pot[i].y = b;

        }

        cnt = ;

        for(int i = ; i <= n- ;i ++){

          for(int j = i +  ; j <= n ; j++){

            rod[cnt].st = i;

            rod[cnt].ed = j;

            rod[cnt++].v = (pot[i].x - pot[j].x) * (pot[i].x - pot[j].x) + (pot[i].y - pot[j].y) * (pot[i].y - pot[j].y);

          }

        }

        sort(rod,rod+cnt);

        for(int i =  ; i <= n ; i++){

            pre[i] = i;

        }

        read(m);

        for(int i = ; i < m; i++){

          read2(a,b);

          join(a,b);

        }

        kruskal();

}