HDU 1856 More is better
More is better
Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 327680/102400 K (Java/Others)
Total Submission(s): 720 Accepted Submission(s): 264
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.
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 2HintA 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
Recommend
lcy
//1256387 2009-04-10 22:20:57 Wrong Answer 1856 250MS 78532K 804 B C++ Wpl
//1256393 2009-04-10 22:21:44 Accepted 1856 250MS 78532K 880 B C++ Wpl
//改变的并查集,实际上就是kruskal求最小生成树的思想
#include <iostream>
#define MAX 10000002 //因为这个题目内存限制是102400 K ,所以开了这么大的数组
using namespace std;
struct node
{
int parent;
int w;
}data[MAX];
int MM;
void Init()
{
int i;
for(i=1;i<MAX;i++)
{
data[i].parent=-1; //这里利用的是kruskal的思想,这里父亲结点全初始化为-1
data[i].w=1;
}
}
int Find(int x) //这里是用来查找根的,就是当前这个x的根节点
{
int i=x;
while(data[i].parent>0)
i=data[i].parent;
return i;
}
void Merge(int a,int b)
{
if(data[a].w>data[b].w)
{
data[a].w+=data[b].w;
data[b].parent=a;
if(data[a].w>MM) //这里用来统计哪个集合的元素个数最多
MM=data[a].w;
}
else
{
data[b].w+=data[a].w;
data[a].parent=b;
if(data[b].w>MM)
MM=data[b].w;
}
}
int main()
{
int n;
while(scanf("%d",&n)!=EOF)
{
MM=1;
int a,b,x,y;
Init();
while(n--)
{
scanf("%d%d",&a,&b);
x=Find(a);
y=Find(b);
if(x!=y)
Merge(x,y);
}
printf("%d\n",MM);
}
return 0;
}
//1256393 2009-04-10 22:21:44 Accepted 1856 250MS 78532K 880 B C++ Wpl
//改变的并查集,实际上就是kruskal求最小生成树的思想
#include <iostream>
#define MAX 10000002 //因为这个题目内存限制是102400 K ,所以开了这么大的数组
using namespace std;
struct node
{
int parent;
int w;
}data[MAX];
int MM;
void Init()
{
int i;
for(i=1;i<MAX;i++)
{
data[i].parent=-1; //这里利用的是kruskal的思想,这里父亲结点全初始化为-1
data[i].w=1;
}
}
int Find(int x) //这里是用来查找根的,就是当前这个x的根节点
{
int i=x;
while(data[i].parent>0)
i=data[i].parent;
return i;
}
void Merge(int a,int b)
{
if(data[a].w>data[b].w)
{
data[a].w+=data[b].w;
data[b].parent=a;
if(data[a].w>MM) //这里用来统计哪个集合的元素个数最多
MM=data[a].w;
}
else
{
data[b].w+=data[a].w;
data[a].parent=b;
if(data[b].w>MM)
MM=data[b].w;
}
}
int main()
{
int n;
while(scanf("%d",&n)!=EOF)
{
MM=1;
int a,b,x,y;
Init();
while(n--)
{
scanf("%d%d",&a,&b);
x=Find(a);
y=Find(b);
if(x!=y)
Merge(x,y);
}
printf("%d\n",MM);
}
return 0;
}
