POJ 1838 Banana (并查集)

Banana

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 2032		Accepted: 755

Description

Consider a tropical forrest, represented as a matrix. The cell from the right top corner of the matrix has the coordinates (1,1), and the coordinates of the other cells are determinated by the row and the column on which the cell is. In some cells of the matrix are placed banana trees; a cell can contain no more than a banana tree. More banana trees which are neighbours on horizontal or vertical form a region of banana trees. In this kind of region, monkey CEKILI is moving easily, with her well-known agility, from a banana tree to another. 
CEKILI is eager and the bananas from a single region are not enough for her. Tarzan wants to help his friend. For that, he may connect exactly k banana tree regions knoting more lianas and so CEKILI could move from a region to another using lianas. Obviously, Tarzan must choose the regions so that the total number of banana trees from those k regions must be maximum. 

Detemine maximum number of banana trees which Tarzan can obtain connecting exactly k regions. 

Input

The input has the following structure: 
Nr K 
x(1) y(1) 
y(2) y(2) 
... 
x(Nr) y(Nr) 
Nr is the number of banana trees. K is the number of zones which can be connected. x(i) is the row of the i-th banana tree, while y(i) is the column of the i-th banana tree. 
There are Constraints: 
• 1 <= Nr <= 16000; 
• 1 <= x(i), y(i) <= 10000; 
• In the tests used for grading k will never be bigger than the number of regions; 
• Two positions are horizontally neighbours if they are on the same row and consecutive columns, respectively vertically neighbours if they are on the same column and on consecutive rows. 

Output

The output will contain on the first line the maximum number of banana trees that can be obtained by connecting the k regions.

Sample Input

10 3
7 10
1 1
101 1
2 2
102 1
7 11
200 202
2 1
3 2
103 1

Sample Output

9

Source

Romania OI 2002

 

题目的意思是有一只猴子想吃香蕉，但是它只能在相邻的树上移动，这里相邻的意思是：横坐标相等时，纵坐标相差1。纵坐标相等时，横坐标相差1。如果两棵树相邻则可以合并到一个区域里面。

问给定n棵树和k个区域，k个区域里面最多有多少香蕉树。

 

看到区域，就想到了并查集。模拟样例之后发现，这里合并挺简单，就是给每个给定的坐标编号，如果符合合并的条件就利用编号将这两棵树合并起来。

#include<cstdio>
#include<iostream>
#include<cstring>
#include<stdlib.h>
#include<algorithm>
using namespace std;
const int MAXN=16000+10;
struct node
{
    int x,y;
    int index;
}a[MAXN];
int p[MAXN],num[MAXN];
int n,k;
int cmp1(const node &A,const node &B)
{
    if(A.x!=B.x)
        return A.x<B.x;
    else
        return A.y<B.y;
}

int cmp2(const node &A,const node &B)
{
    if(A.y!=B.y)
        return A.y<B.y;
    else
        return A.x<B.x;
}

int cmp3(const int a,const int b)
{
    return a>b;
}
int findfa(int x)
{
    return p[x]==x?x:p[x]=findfa(p[x]);
}

int main()
{
    //freopen("in.txt","r",stdin);
    while(scanf("%d %d",&n,&k)!=EOF)
    {
        for(int i=0;i<n;i++)
        {
            scanf("%d %d",&a[i].x,&a[i].y);
            a[i].index=i;
        }

        for(int i=0;i<n;i++)
        {
            p[i]=i;
            num[i]=1;
        }
        sort(a,a+n,cmp1);
        for(int i=0;i<n-1;i++)
        {
            if(a[i].x==a[i+1].x && a[i+1].y-a[i].y==1)
            {
                int xx=findfa(a[i].index);
                int yy=findfa(a[i+1].index);
                if(xx==yy)
                    continue;
                else
                {
                    p[yy]=xx;
                    num[xx]+=num[yy];
                    num[yy]=0;
                }
            }
        }

        sort(a,a+n,cmp2);
        for(int i=0;i<n-1;i++)
        {
            if(a[i+1].x-a[i].x==1 && a[i+1].y==a[i].y)
            {
                int xx=findfa(a[i].index);
                int yy=findfa(a[i+1].index);
                if(xx==yy)
                    continue;
                else
                {
                    p[yy]=xx;
                    num[xx]+=num[yy];
                    num[yy]=0;
                }
            }
        }

        sort(num,num+n);
        int ans=0;
        for(int i=n-1;i>=n-k;i--)
            ans+=num[i];
        printf("%d\n",ans);
    }
    return 0;
}