【杭电1213】How Many Tables

How Many Tables

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

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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

已知会告诉你各个独立集的关系，通过这些关系的作用之后，会留下多少个独立的集合！根据已知条件我们可以讲有关系的集合进行合并，最后看谁的祖先是本身，就说明它将是一个独立集！

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#include<stdio.h>#include<algorithm>using namespace std;int f[1100];int findroot(int x){int r=x;while(r!=f[r])r=f[r];int i=x,j;while(i!=r){j=f[i];f[i]=r;i=j;}return r;}int main(){int t,b,a,n,m;scanf("%d",&t);while(t--){scanf("%d%d",&n,&m);for(int i=1;i<=n;i++)f[i]=i;for(int i=0;i<m;i++){scanf("%d%d",&a,&b);//a与b认识，如果根节点不同，合并两个集合，即成为一桌int xx=findroot(a);int yy=findroot(b);if(xx!=yy)f[xx]=yy;}int num=0;for(int j=1;j<=n;j++){if(f[j]==j)num++;}printf("%d\n",num);}return 0;}
与1232题畅通工程雷同，只需稍作修改

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#include<stdio.h>#include<algorithm>using namespace std;int f[1100];int find(int x){int r=x;while(r!=f[r])r=f[r];int i=x,j;while(i!=r){j=f[i];f[i]=r;i=j;}return r;}void unio(int a,int b){int nx=find(a);int ny=find(b);if(nx!=ny)f[ny]=nx;}int main(){int n,m,a,b,k;scanf("%d",&k);while(k--){scanf("%d%d",&n,&m);for(int i=1;i<=n;i++)f[i]=i;for(int i=0;i<m;i++){scanf("%d%d",&a,&b);unio(a,b);}int t=0;for(int i=1;i<=n;i++){if(f[i]==i)t++;}printf("%d\n",t);}return 0;}