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;
}