HDU1213：How Many Tables

Problem DescriptionToday 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. 

InputThe 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. 

OutputFor 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

 

並查集的基礎題

我擦

我的find函數如果寫成

int find(int x){    int r = x;    while(father[r] != r)        r = father[r];    return r;}

就會逾時，誰跟我說這個函數用時還少些的，我下面的代碼15ms，用了這個就逾時了

#include <stdio.h>int father[1005];int find(int x){    int next;    if(father[x]==x)    return x;    return father[x]=find(father[x]);}void Union(int x,int y){    father[x] = y;}int main(){    int n;    int a,b,i,j;    int x,y,fx,fy;    int ans;    scanf("%d",&n);    while(n--)    {        scanf("%d%d",&a,&b);        for(i = 1; i<=a; i++)            father[i] = i;        while(b--)        {            scanf("%d%d",&x,&y);            fx = find(x),fy = find(y);            if(fx!=fy)            Union(fx,fy);        }        ans = 0;        for(i = 1; i<=a; i++)            if(father[i] == i)                ans++;        printf("%d\n",ans);    }    return 0;}