【POJ 1838 --- Banana】

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

解题思路

分别通过对x,y的排序得到满足相邻的树的坐标。运用并查集标记每一块区域。最后得到每个区域中树的个数，奖其中最多的几个相连。即可解题。

AC代码：

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstdlib>
#include <cstring>
using namespace std;
#define MAXN 16010
struct Node
{int x,y;int num;
}arr[MAXN];
int b[MAXN],result[MAXN];bool compareX(Node &a,Node &b)
{if(a.x != b.x){return a.x < b.x;}return a.y < b.y;
}bool compareY(Node &a,Node &b)
{if(a.y!=b.y){return a.y<b.y;}return a.x < b.x;
}bool compare(int a,int b)
{return a > b;
}
//查找
int seek(int x)
{if(x==b[x])return x;return b[x]=seek(b[x]);
}
//设置连接
int setting(int x,int y)
{x=seek(x);y=seek(y);if(x==y)return 0;elseb[x]=y;return 0;
}int main()
{int n,m,ans=0;scanf("%d%d",&n,&m);for(int i=0;i<n;i++){b[i]=i;scanf("%d%d",&arr[i].x,&arr[i].y);arr[i].num = i+1;}sort(arr,arr+n,compareX);for(int i=0;i<n-1;i++){if(arr[i].x==arr[i+1].x && arr[i+1].y-arr[i].y==1){setting(arr[i].num,arr[i+1].num);}}sort(arr,arr+n,compareY);for(int i=0;i<n-1;i++){if(arr[i].y==arr[i+1].y && arr[i+1].x-arr[i].x==1){setting(arr[i].num,arr[i+1].num);}}//将每个树分别标记到每块区域中for(int i=0;i<n;i++){b[i]=seek(i);}//得到每块区域树的个数for(int i=0;i<n;i++){result[b[i]]++;}sort(result,result+n+1,compare);//获取树的数量最多的m个区域中树的总数for(int i=0;i<m;i++){ans += result[i];}printf("%d\n",ans);return 0;
}

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【POJ 1838 --- Banana】

【POJ 1838 --- Banana】

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