How Many TablesTime Limit: 2000/1000 MS (Java/Others)Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 17994Accepted Submission(s): 8853
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
题意：
有t组数据，每组数据开头为m,n。表示有m个人，下面n行表示a,b相互认识，认识可以传递，即a认识b,b认识c，那么a就认识c。认识的人可以用同一张桌子，问一共需要多少张桌子。
题解：
将认识的人归为一组，可以用并查集解决。
参考代码：
#include<stdio.h>#include<string.h>#define N 1005int father[N];int f(int n){return father[n]==n?n:father[n]=f(father[n]);} void merge(int x,int y){int fx,fy;fx=f(x);fy=f(y);if(fx!=fy)father[fx]=fy;}int main(){int t;scanf("%d",&t);while(t–){int m,n,a,b,sum;sum=0;for(int i=1;i<N;i++)father[i]=i;scanf("%d%d",&m,&n);for(int i=0;i<n;i++){scanf("%d%d",&a,&b);merge(a,b);}for(int i=1;i<=m;i++){if(father[i]==i)sum++;}printf("%d\n",sum);}return 0;}
版权声明：本文为博主原创文章，，随便转载。
接受失败等于打破完美的面具，接受失败等于放松自己高压的心理，
