HDOJ--1213--How Many Tables【并查集】

Problem 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

 

Author

Ignatius.L

 

Source

杭电ACM省赛集训队选拔赛之热身赛

 

Recommend

Eddy

 

思路：并查集的简单题，多定义一个s，当两个人可以坐同一张桌子时s+1，表示多少人可以不用单独一桌，最后用总人数n减去s就是所需的桌子数。

#include<stdio.h>#include<cstring>#include<algorithm>#include<math.h>#include<iostream>#include<time.h>using namespace std;int father[1005];int s;int find(int x){    if(father[x]!=x) father[x]=find(father[x]);    return father[x];}void combine(int x,int y){    int a,b;    a=find(x);    b=find(y);    if(a!=b)    {        father[a]=b;        s++;    }}int main(){    int t,n,m,x,y,i;    scanf("%d",&t);    while(t--)    {        s=0;        scanf("%d%d",&n,&m);        for(i=1;i<=n;i++)            father[i]=i;        for(i=0;i<m;i++)        {            scanf("%d%d",&x,&y);            combine(x,y);        }        printf("%d\n",n-s);    }    return 0;}