hdu 1856 More is better （并查集）

Problem Description
Mr Wang wants some boys to help him with a project. Because the project is rather complex, the more boys come, the better it will be. Of course there are certain requirements.

Mr Wang selected a room big enough to hold the boys. The boy who are not been chosen has to leave the room immediately. There are 10000000 boys in the room numbered from 1 to 10000000 at the very beginning. After Mr Wang’s selection any two of them who are still in this room should be friends (direct or indirect), or there is only one boy left. Given all the direct friend-pairs, you should decide the best way.

Input
The first line of the input contains an integer n (0 ≤ n ≤ 100 000) - the number of direct friend-pairs. The following n lines each contains a pair of numbers A and B separated by a single space that suggests A and B are direct friends. (A ≠ B, 1 ≤ A, B ≤ 10000000)

Output
The output in one line contains exactly one integer equals to the maximum number of boys Mr Wang may keep.

Sample Input
4
1 2
3 4
5 6
1 6
4
1 2
3 4
5 6
7 8
题意：给出n，接下来n行表示a,b之间连通，求出最大的联通块数量，注意n为0是输出1；
思路：用一个数组记录连通块的数量，在合并时也把数量记下；

代码：

#include<cstdio>#include<algorithm>#define maxn 10000005using namespace std;int a[maxn],vag[maxn];int maxz;int find(int t){    if(t!=a[t])       a[t]=find(a[t]);    return a[t];}void mark(int x,int y){    int zx=find(x);    int zy=find(y);    if(zx==zy)       return;    if(vag[zx]<vag[zy])    {        a[zx]=zy;        vag[zy]+=vag[zx];        vag[zx]=0;        if(maxz<vag[zy])           maxz=vag[zy];    }    else    {        a[zy]=x;        vag[zx]+=vag[zy];        vag[zy]=0;        if(maxz<vag[zx])           maxz=vag[zx];    }}int main(){    int t;    while(~scanf("%d",&t))    {        int n,m;        if(t==0)        {            printf("1\n");            continue;        }        maxz=0;        for(int i=1;i<=maxn;i++)        {            a[i]=i;            vag[i]=1;        }        for(int i=1;i<=t;i++)        {            int za,zb;            scanf("%d%d",&za,&zb);            mark(za,zb);        }        printf("%d\n",maxz);    }}