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
42HintA 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. Author
lxlcrystal@TJU
Source
思路
裸并查集，只要找到所有分出来的集合当中元素最多的集合就行了，为此需要一个数组来记录个数，详见代码
代码
#includeusing namespace std;int father[10000010];int num[10000010];int MaxValue;void init(int n){ for(int i=1;i<=n;i++) { father[i]=i; num[i] = 1; }}int find(int x){ if(father[x]!=x) father[x] = find(father[x]); return father[x];}void join(int a,int b){ int t1=find(a); int t2=find(b); if(t1!=t2) { father[t1]=t2; num[t2] += num[t1]; MaxValue = max(num[t2],MaxValue); }}int main(){ int n; while(scanf("%d",&n)!=EOF) { if(n==0) { cout << 1 << endl; continue; } init(10000000); MaxValue = -1; for(int i=1;i<=n;i++) { int a,b; scanf("%d%d",&a,&b); if(find(a) != find(b)) join(a,b); } cout << MaxValue << endl; } return 0;}