题目描述
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.
输入
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)
输出
The output in one line contains exactly one integer equals to the maximum number of boys Mr Wang may keep.
样例输入
3
1 3
1 5
2 5
4
3 2
3 4
1 6
2 6样例输出
4
5题目大意： 房间里有编号从1~10000000共10000000人，每次给出n对关系，每对关系表示这两个人被选中了且成为朋友。问最后被选中的最大朋友人数是多少，如果没有人被选中，则留下1个人；如果有多组朋友，输出最大的组有多少人。
分析：并查集的应用。不过这个集合很大，因此要在合并的时候，记录合并后组的人数。
#include<algorithm>
#include <iostream>
#include <cstdlib>
#include <cstring>
#include <string>
#include <vector>
#include <cstdio>
#include <queue>
#include <stack>
#include <ctime>
#include <cmath>
#include <map>
#include <set>
#define INF 0xffffffff
#define db1(x) cout<<#x<<"="<<(x)<<endl
#define db2(x,y) cout<<#x<<"="<<(x)<<", "<<#y<<"="<<(y)<<endl
#define db3(x,y,z) cout<<#x<<"="<<(x)<<", "<<#y<<"="<<(y)<<", "<<#z<<"="<<(z)<<endl
#define db4(x,y,z,r) cout<<#x<<"="<<(x)<<", "<<#y<<"="<<(y)<<", "<<#z<<"="<<(z)<<", "<<#r<<"="<<(r)<<endl
#define db5(x,y,z,r,w) cout<<#x<<"="<<(x)<<", "<<#y<<"="<<(y)<<", "<<#z<<"="<<(z)<<", "<<#r<<"="<<(r)<<", "<<#w<<"="<<(w)<<endl
using namespace std;int father[10000005],cnt[10000005];int findFather(int x,int father[])
{int a=x;while(father[x]!=x)x=father[x];while(a!=father[a]){int temp=a;a=father[a],father[temp]=x;}return x;
}void Union(int a,int b,int father[],int cnt[],int &ans)
{int faA=findFather(a,father),faB=findFather(b,father);
// db4(a,b,faA,faB);if(faA!=faB){father[faA]=faB;cnt[faB]+=cnt[faA];ans=max(ans,cnt[faB]);}return;
}int main(void)
{#ifdef testfreopen("in.txt","r",stdin);//freopen("in.txt","w",stdout);clock_t start=clock();#endif //testint T,n,m;while(~scanf("%d",&T)){int ans=-1;if(T==0){printf("1\n");continue;}for(int i=1;i<=10000005;++i)father[i]=i,cnt[i]=1;for(int i=0;i<T;++i){scanf("%d%d",&n,&m);Union(n,m,father,cnt,ans);}printf("%d\n",ans);}#ifdef testclockid_t end=clock();double endtime=(double)(end-start)/CLOCKS_PER_SEC;printf("\n\n\n\n\n");cout<<"Total time:"<<endtime<<"s"<<endl; //s为单位cout<<"Total time:"<<endtime*1000<<"ms"<<endl; //ms为单位#endif //testreturn 0;
}
