HDOJ 1213 How Many Tables 1232 畅通工程（并查集基础，水）

How Many Tables

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 28401    Accepted Submission(s): 14052

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

题目比较简单，直接套并查集，查询集合数，我的代码还不够优化

#include <iostream>using namespace std;const int MAX_N=1005;int par[MAX_N];int t,x,y,m,n;void init(int n)//初始化{    for(int i=0;i<=n;i++)    {        par[i]=i;    }}int find(int x){    return par[x]==x?x:(par[x]=find(par[x]));//路径压缩}void unit(int x,int y){    x=find(x);    y=find(y);    if(x==y)    return;    else    par[x]=y;}/*bool same(int x,int y){    return find(x)==find(y);}*/int main(){    cin>>t;    while(t--)    {        cin>>n>>m;        init(n);        for(int i=0;i<m;i++)        {            cin>>x>>y;            unit(x,y);        }        int sum=0;        for(int i=1;i<=n;i++)        {            if(par[i]==i)                sum++;        }        cout<<sum<<endl;    }    return 0;}

1232可参考1213，一样的思路