# 分治算法实例代码

目录

• 平面点集最接近点对问题
• 选择问题
• 残缺棋盘问题

平面点集最接近点对问题

大致是要先横坐标排序，再纵坐标排序。
OI-wiki详解

选择问题

和快速排序的过程有点相似，都需要将元素分成小于某个数的一堆，和大于这个数的一堆，然后递归下去。

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<algorithm>
using namespace std;
int a[1001];

int select(int left,int right,int k){
    if( left >= right )
        return a[left];
    int x = a[left];
    int i = left;
    int j = right+1;
    while( true ){
        do{
            i++;
        }while(a[i]<x);
        do{
            j--;
        }while(a[j]>x);
        if( i>=j )
            break ;
        swap(a[i],a[j]);
    }
    if( j-left+1 == k )
        return x;
    a[left] = a[j];
    a[j] = x;
    if( j-left+1 < k )
        return select(j+1,right,k-j+left-1);//左闭右闭 区间
    else
        return select(left,j-1,k);
}

int main(){
    int n;
    while( scanf("%d",&n) != EOF ){
        int k;//第k小
        cin>>k;
        for( int i = 0; i < n; i++ )
            cin>>a[i];
        cout<<select(0,n-1,k)<<endl;
    }
    return 0;
}

残缺棋盘问题

详解

#include <iostream>
 
using namespace std;
 
int amount,Board[100][100];
 
void Cover(int tr,int tc,int dr,int dc,int size)
{
    int s,t;
    if(size<2)
        return ;
    amount++;
    t=amount;
    s=size/2;
    if(dr<tr+s&&dc<tc+s)//残缺在左上角
    {
        //覆盖中间位置
        Board[tr+s-1][tc+s]=t;
        Board[tr+s][tc+s-1]=t;
        Board[tr+s][tc+s]=t;
 
        Cover(tr,tc,dr,dc,s);//覆盖左上
        Cover(tr,tc+s,tr+s-1,tc+s,s);//覆盖右上
        Cover(tr+s,tc,tr+s,tc+s-1,s);//覆盖左下
        Cover(tr+s,tc+s,tr+s,tc+s,s);//覆盖右下
    }
    else if(dr<tr+s&&dc>=tc+s)//残缺在右上角
    {
        Board[tr+s-1][tc+s-1]=t;
        Board[tr+s][tc+s-1]=t;
        Board[tr+s][tc+s]=t;
 
        Cover(tr,tc,tr+s-1,tc+s-1,s);
        Cover(tr,tc+s,dr,dc,s);
        Cover(tr+s,tc,tr+s,tc+s-1,s);
        Cover(tr+s,tc+s,tr+s,tc+s,s);
    }
    else if(dr>=tr+s&&dc<tc+s)//残缺在左下
    {
        Board[tr+s-1][tc+s-1]=t;
        Board[tr+s-1][tc+s]=t;
        Board[tr+s][tc+s]=t;
 
        Cover(tr,tc,tr+s-1,tc+s-1,s);
        Cover(tr,tc+s,tr+s-1,tc+s,s);
        Cover(tr+s,tc,dr,dc,s);
        Cover(tr+s,tc+s,tr+s,tc+s,s);
    }
    else
    {
        Board[tr+s-1][tc+s-1]=t;
        Board[tr+s-1][tc+s]=t;
        Board[tr+s][tc+s-1]=t;
 
        Cover(tr,tc,tr+s-1,tc+s-1,s);
        Cover(tr,tc+s,tr+s-1,tc+s,s);
        Cover(tr+s,tc,tr+s,tc+s-1,s);
        Cover(tr+s,tc+s,dr,dc,s);
    }
}
 
void Print(int s)
{
    for(int i=1;i<=s;i++)
    {
        for(int j=1;j<=s;j++)
            printf("%5d ",Board[i][j]);
        printf("\n");
    }
}
 
int main()
{
    int s=1,k,x,y;
    printf("输入2残缺棋盘的规模:2^k,k=");
    scanf("%d",&k);
    for(int i=1;i<=k;i++)
        s*=2;
    printf("输入棋盘残缺位置(x,y):");
    scanf("%d%d",&x,&y);
    Board[x][y]=0;
    Cover(1,1,x,y,s);
    Print(s);
    return 0;
}