Little Bishops uva861

最新推荐文章于 2020-10-22 00:09:19 发布

转载最新推荐文章于 2020-10-22 00:09:19 发布 · 128 阅读

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原文链接：http://www.cnblogs.com/767355675hutaishi/p/3829594.html

本文介绍了一个算法问题：如何在n×n的棋盘上放置k个象，使得任意两个象不在互相攻击的位置。文章给出了具体的实现代码，包括初始化棋盘、计算不同颜色方格上象的放置方式等步骤。

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Little Bishops

A bishop is a piece used in the game of chess which is played on a board of square grids. A bishop can only move diagonally from its current position and two bishops attack each other if one is on the path of the other. In the following figure, the dark squares represent the reachable locations for bishop B₁ form its current position. The figure also shows that the bishops B₁ and B₂ are in attacking positions whereas B₁ andB₃ are not. B₂ and B₃ are also in non-attacking positions.

Now, given two numbers n and k, your job is to determine the number of ways one can put k bishops on an n × n chessboard so that no two of them are in attacking positions.

Input

The input file may contain multiple test cases. Each test case occupies a single line in the input file and contains two integers n (1 ≤ n ≤ 8) and k(0 ≤ k ≤ n²).

A test case containing two zeros for n and k terminates the input and you won’t need to process this particular input.

Output

For each test case i

#include"iostream"
#include"cstring"
#include"algorithm"
using namespace std;
const int N=8;
int b[N+1],w[N+1],rb[N+1][65],rw[N+1][65];
void init_chessboard(int n)
{
    memset(b,0,sizeof(b));
    memset(w,0,sizeof(w));
    for(int i=1;i<=n;i++)
      for(int j=1;j<=n;j++)
    {
        if((i+j)&1)
           w[(i+j)>>1]++;
        else
           b[(i+j)>>1]++;
    } 
}
void bishops(int n,int k,int c[N+1],int R[N+1][65])
{
    for(int i=0;i<=n;i++)
      R[i][0]=1;
    for(int j=1;j<=k;j++)
        R[0][j]=0;
    for(int i=1;i<=n;i++)
      for(int j=1;j<=c[i];j++)
        R[i][j]=R[i-1][j]+R[i-1][j-1]*(c[i]-j+1);
}
int main()
{
    int n,k,ans;
    while(scanf("%d%d",&n,&k)!=EOF)
    {
        if(n==0&&k==0)
          break;
        init_chessboard(n);
        sort(b+1,b+n+1);
        sort(w+1,w+n);
        bishops(n,k,b,rb);
        bishops(n-1,k,w,rw);
        ans=0;
        for(int i=0;i<=k;i++)
          ans+=rb[n][i]*rw[n-1][k-i];
        printf("%d\n",ans);
    }
    return 0;
}

n the input print a line containing the total number of ways one can put the given number of bishops on a chessboard of the given size so that no two of them are in attacking positions. You may safely assume that this number will be less than 10¹⁵.

Sample Input
8 6
4 4
0 0

Sample Output
5599888
260

转载于:https://www.cnblogs.com/767355675hutaishi/p/3829594.html