Uva 861 (little bishop)搜索，棋盘多项式，dp

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[b]Little Bishops

[/b]



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

0

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

there are some good example

input:

8 6

4 4

8 15

8 12

8 0

5 1

5 5

5 9

8 10

1 1

1 0

7 8

0 0

out put

5599888

260

0

489536

0

25

3368

0

12448832

1

0

867328

#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
[k]<<endl;
    }
return 0;
}

other excellent solutions

1.棋盘多项式：http://blog.tianya.cn/blogger/post_show.asp?idWriter=0&Key=0&BlogID=397009&PostID=6849511

2.dp：http://www.cppblog.com/lzh/archive/2010/08/25/124684.html