hdoj_1856More is better（并查集）

More is better

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 327680/102400 K (Java/Others)
Total Submission(s): 8846    Accepted Submission(s): 3303

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

标记个数组，记录节点的个数！

#include<iostream>#include<cstdio>#include<map>#include<cstring>#include<cmath>#include<vector>#include<algorithm>#include<set>#include<string>#include<queue>#include <stack>using namespace std;#pragma warning(disable : 4996)const int MAX = 10000005;int father[MAX];int ranks[MAX];/* 初始化集合 */void Make_Set(int n){for(int x = 1; x <= n; x++){father[x] = x;ranks[x] = 1;}}/* 查找x元素所在的集合,回溯时压缩路径 */int Find_Set(int x){if (x != father[x]){father[x] = Find_Set(father[x]);}return father[x];}void Union_Set(int x, int y){x = Find_Set(x);y = Find_Set(y);if (x == y) return;father[x] = y;ranks[y] = ranks[x] + ranks[y];}int main(){freopen("in.txt", "r", stdin);int n, x, y;while (scanf("%d", &n) != EOF){Make_Set(MAX);while (n--){scanf("%d %d", &x, &y);Union_Set(x, y);}int ans = -1;for(int i = 1; i <= MAX; i++){if(ranks[i] > ans){ans = ranks[i];}}cout << ans << endl;}return 0;}