HDU 1856 More is better（并查集）
题目：
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.
InputThe 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)OutputThe 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
Sample Output
4 2
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.
大意：
王先生想让一些男孩帮他做一个项目。因为这个项目相当复杂，男孩越多，情况就越好。当然也有一定的要求。
王先生选了一个大得足以容纳男孩的房间。未被选中的男孩必须立即离开房间。房间里有10000000个男孩，从一开始就从1到10000000。
在王先生的选择之后，他们中还有两个人仍然是朋友（直接或间接），或者只剩下一个男孩。给定所有的直接朋友对，你应该决定最好的方式
分析：
假设A和B是朋友，B和C是朋友，那么A和C也是朋友（间接），以此类推，找出最大的朋友数，并查集加一个数组标记；
附代码；
1 #include<map> 2 #include<cmath> 3 #include<queue> 4 #include<vector> 5 #include<cstdio> 6 #include<cstring> 7 #include<iostream> 8 #include<algorithm> 9 using namespace std; 10 int p[1000005],r[1000005]; 11 int find(int x) 12 { 13 int r=x; 14 while (p[r]!=r) 15 r=p[r ]; 16 int j,i=x; 17 while(i!=r) 18 { 19 j=p[i ]; 20 p[i ]=r; 21 i=j; 22 } 23 return r; 24 } 25 int main() 26 { 27 int n; 28 while(~scanf("%d",&n)) 29 { 30 if(n==0) 31 { 32 printf("1\n"); 33 continue; 34 } 35 for(int i=1; i<1000005; i++) 36 { 37 p[i]=i; 38 r[i]=1; 39 } 40 int all=0; 41 for(int i=0; i<n; i++) 42 { 43 int f1,f2,p1,p2; 44 scanf("%d%d",&p1,&p2); 45 f1=find(p1); 46 f2=find(p2); 47 if(f1!=f2) 48 { 49 p[f1]=f2; 50 r[f2]+=r[f1]; 51 } 52 if(all<r[p[p2]]) 53 all=r[p[p2]]; 54 } 55 printf("%d\n",all); 56 } 57 return 0; 58 }
第一次写博客，请多指教！！！
