Uva 861 (little bishop)搜索，棋盘多项式，dp

最新推荐文章于 2018-09-17 19:45:49 发布

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本文探讨了在n×n的国际象棋棋盘上放置k个象，使得任意两个象都不处于相互攻击的位置的问题。提供了两种解决方案，一种是通过递归搜索的方法，另一种则是利用预计算表格的方式大幅提高效率。

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Little Bishops

A bishop is a piece used in thegame of chess which is played on a board of square grids. A bishop can only movediagonally from its current position and two bishops attack each other if oneis on the path of the other. In the following figure, the dark squaresrepresent the reachable locations for bishop B₁ formits current position. The figure also shows that the bishopsB₁and B₂ are in attacking positions whereasB₁and B₃ are not.B₂ and B₃are also in non-attacking positions.

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

analysis: the moment I read the problem, an idea of dfs occurred to me. so I tryed it .It turned out to be a result of TL!!!

For the n is no more than 8,I decided to use array to store all the possible answers.

the former code

later, I got the point why I got time out

e.g.

in put

8 15

output

the example will cost you a few seconds (surely more the 2).T ^T

there are some good example

input:

out put

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
using namespace std;


long k, n, ans,vis[3][60],map[20][20];

void solve ( long dep, long x, long y )
{

    if ( dep == k )
    {
        ans ++;//printf("aaa\n");
        return;
    }
    for(long i=x; i<=n; i++)
    {
        int tmp=1;
        if(i==x) tmp=y+1;
        for(long j=tmp; j<=n; j++)
        {
            if(!map[i][j]&&!vis[0][i-j+n]&&!vis[1][i+j])
            {
                map[i][j]=vis[0][i-j+n]=vis[1][i+j]=1;
                solve(dep+1,i,j);
                map[i][j]=vis[0][i-j+n]=vis[1][i+j]=0;

            }
        }
    }


}

int main()
{
    long i,j;
    while (1 )
    {
        scanf ( "%ld %ld", &n, &k );
        if (!n&&!k) break;
        ans = 0;

        memset ( map, 0, sizeof(map));
        memset ( vis, 0, sizeof(vis) );
        for(i=1; i<=n; i++)
            for(j=1; j<=n; j++)
            {
                map[i][j]=vis[0][i-j+n]=vis[1][i+j]=1;
                solve ( 1, i, j );
                map[i][j]=vis[0][i-j+n]=vis[1][i+j]=0;

            }
        printf ( "%ld\n", ans );


    }
    return 0;
}

the latter code

#include<iostream>
using namespace std;

const long cheat[9][67]={{0},
                         {1,1},
                         {1,4,4},
                         {1,9,26,26,8},
                         {1,16,92,232,260,112,16},
                         {1,25,240,1124,2728,3368,1960,440,32},
                         {1,36,520,3896,16428,39680,53744,38368,12944,1600,64},
                         {1,49,994,10894,70792,282248,692320,1022320,867328,389312,81184,5792,128},
                         {1,64,1736,26192,242856,1444928,5599888,14082528,22522960,22057472,12448832,3672448,489536,20224,256}};

int main()
{
    long n,k;
    
    while(cin>>n>>k)
    {
       if(n==0&&k==0) break;
       
       cout<<cheat[n][k]<<endl;
    }
return 0;
}