Posted on 2010-08-10 15:03
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ACM ( 并查集 )
MiYu原創(chuàng), 轉(zhuǎn)帖請注明 : 轉(zhuǎn)載自 ______________白白の屋
題目地址:
http://acm.hdu.edu.cn/showproblem.php?pid=1856
題目描述:
More is better
Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 327680/102400 K (Java/Others)
Total Submission(s): 1710 Accepted Submission(s): 643
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
4
1 2
3 4
5 6
1 6
4
1 2
3 4
5 6
7 8
Sample Output
4
2
題目分析:
如果對并查集比較熟習(xí)的話, 這道題就可以直接模板AC了. 不了解的話請點(diǎn)擊 :
并查集 學(xué)習(xí) 詳解 這道題目的意思就是在 所有選出的集合中選出最大的集合, 也就是人最多的集合, 另外, 如果所有點(diǎn)
都是孤立點(diǎn), 也就是說所有人都互不認(rèn)識(shí), 那么 答案顯然是 1 了.
代碼如下 :
MiYu原創(chuàng), 轉(zhuǎn)帖請注明 : 轉(zhuǎn)載自 ______________白白の屋
#include <iostream>
using namespace std;
typedef struct {
int parent;
int cnt;
}Tset;
int maxSet = 0;
typedef struct treeUFS{
public:
treeUFS(int n = 0):N(n+1) { set = new Tset[N];
for ( int i = 0; i != N; ++ i)
set[i].parent = i,set[i].cnt = 1;
}
~treeUFS(){ delete [] set; };
int find ( int x ){ int r = x; while ( set[r].parent != r )
r = set[r].parent;
int i = x;
while ( i != r) {
int j = set[i].parent;
set[i].parent = r;
i = j;
}
return r;
}
void init () { for ( int i = 0; i != N; ++ i) set[i].parent = i,set[i].cnt = 1; }
void Merge( int x,int y ){ x = find ( x ); y = find ( y );
if ( x == y ) return;
if ( set[x].cnt > set[y].cnt ){
set[y].parent = x;
set[x].cnt += set[y].cnt;
if ( set[x].cnt > maxSet ){
maxSet = set[x].cnt ;
}
}
else{
set[x].parent = y;
set[y].cnt += set[x].cnt;
if ( set[y].cnt > maxSet ){
maxSet = set[y].cnt ;
}
}
}
int getSetCount ( int x ){ return set[ find(x) ].cnt; }
private:
Tset *set;
int N;
}treeUFS;
int main ()
{
int N,a,b;
treeUFS UFS ( 10000000 );
while ( scanf ( "%d", &N ) != EOF )
{
maxSet = 0;
for ( int i = 1; i <= N; ++ i )
{
scanf ( "%d%d", &a,&b );
UFS.Merge ( a,b );
}
printf ( maxSet == 0 ? "1\n" : "%d\n",maxSet );
UFS.init ();
}
return 0;
}