hdu 1856 More is better(并查集)

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1856

More is better

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 327680/102400 K (Java/Others)
Total Submission(s): 18985    Accepted Submission(s): 6990

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

41 23 45 61 641 23 45 67 8

 

Sample Output

42
Hint
A and B are friends(direct or indirect), B and C are friends(direct or indirect), then A and C are also friends(indirect). In the first sample {1,2,5,6} is the result.In the second sample {1,2},{3,4},{5,6},{7,8} are four kinds of answers. 

一定要仔细看题目啊！！！

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 327680/102400 K (Java/Others)

时间限制虽然在1s 但是内存放的很宽啊！， 所以什么数组存不下等等的顾虑都抛开吧，大胆的开数组！

【代码】

#include <cstdio>const int maxn= 10000000+10;int father[maxn];int num[maxn];int Find(int x){ //路径压缩迭代版本    int root = x;    while(root!=father[root])         root = father[root];//肇东根节点    while(x!=root){ //将这棵树上的节点都指向根节点        int tmp = father[x];        father[x] = root;        x= tmp;    }    return root;}void Union(int a,int b){    int p=Find(a);    int q=Find(b);    if(p!=q){        father[p]=q;        num[q]+=num[p]; //将合并后的数量加到 父节点上    }}int main(){    int n;    while(scanf("%d",&n)!=EOF){        if(n==0){            puts("1");continue;        }        for(int i=1;i<=maxn;i++){            num[i]=1;            father[i]=i;        }        int a,b;        int Max=0;        for(int i=0;i<n;i++){            scanf("%d%d",&a,&b);            if(a>Max)                Max=a;            if(b>Max)                Max=b;            Union(a,b);        }        //printf("sd");        int max=0;         for(int i=1;i<=Max;i++){             if(num[i]>max)                 max=num[i];            // printf("bug\n");         }        printf("%d\n",max);    }    return 0;}