并查集

http://vjudge.net/contest/view.action?cid=49839#problem/A

A - A

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

Submit Status

Appoint description: System Crawler  (2014-03-27)

Description

Today is Ignatius' birthday. He invites a lot of friends. Now it's dinner time. Ignatius wants to know how many tables he needs at least. You have to notice that not all the friends know each other, and all the friends do not want to stay with strangers. 

One important rule for this problem is that if I tell you A knows B, and B knows C, that means A, B, C know each other, so they can stay in one table. 

For example: If I tell you A knows B, B knows C, and D knows E, so A, B, C can stay in one table, and D, E have to stay in the other one. So Ignatius needs 2 tables at least. 

 

Input

The input starts with an integer T(1<=T<=25) which indicate the number of test cases. Then T test cases follow. Each test case starts with two integers N and M(1<=N,M<=1000). N indicates the number of friends, the friends are marked from 1 to N. Then M lines follow. Each line consists of two integers A and B(A!=B), that means friend A and friend B know each other. There will be a blank line between two cases. 

 

Output

For each test case, just output how many tables Ignatius needs at least. Do NOT print any blanks. 

 

Sample Input

 25 31 22 34 55 12 5 

 

Sample Output

 24 

 

加入了压缩路径

#include<stdio.h>  #include<stdlib.h>  #include<algorithm>  #include<iostream>  using namespace std;  int m,n,parent[20000];void UFset()//初始化{for(int i=1;i<=m;i++)parent[i]=-1;}int Find(int x){int s;//查找位置，一直找到parent[s]为负数为止for(s=x;parent[s]>=0;s=parent[s]);while(s!=x){int tmp=parent[x];//优化方案，压缩路径，更新子节点parent[x]=s;x=tmp;}return s;}void Union( int R1,int R2 ){int r1=Find(R1),r2=Find(R2);int tmp=parent[r1]+parent[r2];//两个集合结点个数之和（负数）if(parent[r1]>parent[r2])//优化方案--加权法则{parent[r1]=r2;//根节点r1所在的树作为r2的子树parent[r2]=tmp;}else{parent[r2]=r1;//<span style="font-family: Arial, Helvetica, sans-serif;">//根节点r2所在的树作为r1的子树</span>parent[r1]=tmp;}}int main(){int N;scanf("%d",&N);while(N--){scanf("%d%d",&m,&n);UFset();int a,b;for(int i=0;i<n;i++){scanf("%d%d",&a,&b);if(Find(a) != Find(b))//这里要进行判断父节点不等，才经行合并Union(a,b);}int num=0;for(int i=1;i<=m;i++){if(parent[i]<0)num++;}printf("%d\n",num);}return 0;}