hdu 1856 More is better

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More is better Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 327680/102400 K (Java/Others)
Total Submission(s): 14656    Accepted Submission(s): 5390

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.

 

題意： 給出兩兩關係， 求最大集合的元素個數。

思路：初始化每個集合的cnt為1，合并過程中合并cnt，並更新最大cnt

換了種寫法，p初始化為-1，find_set 時候判斷是否是-1，如果是，說明是根節點。

#include<cstring>#include<iostream>#include<string>#include<algorithm>using namespace std;const int maxn=10000000+5;int p[maxn], cnt[maxn], max_cnt;void make_set(){    memset(p, -1, sizeof(p));    for(int i=0;i<maxn;i++) cnt[i]=1;    max_cnt=1;}int find_set(int x){    return p[x]==-1 ? x : p[x]=find_set(p[x]);    //return p[x]==x ? x : p[x]=find_set(p[x]);}void union_set(int x, int y){    int fx=find_set(x), fy=find_set(y);    if(fx==fy) return;    p[fx]=fy;    cnt[fy]+=cnt[fx];    max_cnt=max(max_cnt, cnt[fy]);}int main(){    int n, x, y;    while(scanf("%d", &n)!=EOF) {        make_set();        for(int i=0;i<n;i++)        {            scanf("%d%d", &x, &y); union_set(x, y);        }        printf("%d\n", max_cnt);    }    return 0;}