URAL 1034 Queens in Peaceful Positions

URAL_1034

    和经典的N皇后问题很相似，尽管N比较大，但由于只重新安排3个皇后有C(N,3)种情况，每个皇后至多有3个位置可以放，所以最大也就是有C(50,3)*3^3种情况。

    位运算版的N皇后问题详见Matrix67的博客：http://www.matrix67.com/blog/archives/266。

#include<stdio.h>

#include<string.h>

#define MAXD 60

int N, cnt, visr[MAXD], visc[MAXD], vis[MAXD][MAXD];

struct Point

{

    int x, y;

}p[MAXD];

void init()

{

    int i;

    memset(vis, 0, sizeof(vis));

    for(i = 0; i < N; i ++)

    {

        scanf("%d%d", &p[i].x, &p[i].y);

        vis[p[i].x][p[i].y] = 1;

    }

}

int precheck(int x, int y, int z)

{

    int i, j, k;

    memset(visr, 0, sizeof(visr));

    memset(visc, 0, sizeof(visc));

    for(i = 0; i < N; i ++)

        if(i != x && i != y && i != z)

        {

            if(visr[p[i].x])

                return 0;

            visr[p[i].x] = p[i].y;

            if(visc[p[i].y])

                return 0;

            visc[p[i].y] = p[i].x;

        }

}

long long cstate()

{

    int i;

    long long cs = 0;

    for(i = 1; i <= N; i ++)

        if(visc[i])

            cs |= 1ll << i;

    return cs;

}

void dfs(int cur, long long cs, long long ls, long long rs)

{

    int y;

    long long st;

    if(cur > N)

    {

        ++ cnt;

        return ;

    }

    if(y = visr[cur])

    {

        st = ls | rs;

        if(st & (1ll << y))

            return ;

        dfs(cur + 1, cs, (ls | (1ll << y)) << 1, (rs | (1ll << y)) >> 1);

    }

    else

    {

        st = cs | ls | rs;

        for(y = 1; y <= N; y ++)

            if((st & (1ll << y)) == 0 && !vis[cur][y])

                dfs(cur + 1, cs | (1ll << y), (ls | (1ll << y)) << 1, (rs | (1ll << y)) >> 1);

    }

}

void solve()

{

    int i, j, k, p;

    cnt = 0;

    for(i = 0; i < N; i ++)

        for(j = i + 1; j < N; j ++)

            for(k = j + 1; k < N; k ++)

                if(precheck(i, j, k))

                    dfs(1, cstate(), 0, 0);

    printf("%d\n", cnt);

}

int main()

{

    while(scanf("%d", &N) == 1)

    {

        init();

        if(N < 4)

            printf("0\n");

        else

            solve();

    }

    return 0;

}