1272. Non-Yekaterinburg Subway

A little town started to construct a subway. The peculiarity of the town is that it is located on small islands, some of them are connected with tunnels or bridges. The mayor is sure that the subway is to be under the ground, that’s why the project must use the less bridges the better. The only request for the subway is that the townsmen could get by metro (may be with changes) from every island to every island. Fortunately, we know that there is enough tunnels and bridges for it. It was decided to construct as less passages from island to island as possible to save money.

Your task given a town plan to determine the minimal possible number of bridges that is necessary to use in the subway construction.

Input

The first line contains three integers separated with a space: N (the number of islands,1 ≤ N ≤ 10000), K (the number of tunnels, 0 ≤ K ≤ 12000) and M (the number of bridges,0 ≤ M ≤ 12000). Then there are K lines; each line consists of two integers — the numbers of islands, connected with the corresponding tunnel. The last M lines define bridges in the same format.

Output

the minimal number of bridges necessary for the subway construction.

Sample

input	output
6 3 4 1 2 2 3 4 5 1 3 3 4 4 6 5 6	2

Problem Author: Magaz Asanov (prepared Igor Goldberg) Problem Source: Ural State University championship, October 25, 2003

Tags: (

show tags for all problems

)

Difficulty: 202    Printable version    Submit solution    Discussion (19) My submissions    All submissions (16757)    All accepted submissions (5492)   Solutions rating (1737)

**********************************************************************************************************

并查集，找最小联通

*********************************************************************************************************

#include<iostream>
#include<string>
#include<cstring>
#include<cmath>
#include<cctype>
#include<cstdio>
#include<stack>
#include<queue>
#include<vector>
using namespace std;
int   fa[10010];
int n,m,k,i,j;
struct node
{
    int s,e,c;
}nodea[25000];
int father(int x)
 {
     if(fa[x]!=x)
       fa[x]=father(fa[x]);
     return fa[x];
 }
 int main()
 {
     cin>>n;
     cin>>m>>k;
     for(i=1;i<=m;i++)
      {
          cin>>nodea[i].s>>nodea[i].e;
          nodea[i].c=0;
      }
     for(i=1;i<=k;i++)
      {
          cin>>nodea[i+m].s>>nodea[i+m].e;
          nodea[i+m].c=1;
      }
      int sum=0,l=0;
       for(i=1;i<=n;i++)
        fa[i]=i;
      for(i=1;i<=k+m;i++)
       {
          int x1=father(nodea[i].s);
          int x2=father(nodea[i].e);
          if(x1!=x2)
            {
                sum+=nodea[i].c;
                l++;
                if(l==n-1)break;
                fa[x1]=x2;
            }

       }
     cout<<sum<<endl;
     return 0;
 }