SGU 222 Little Rooks

SGU_222

最近刚学了插头dp，所以就用插头dp那个模式写了这个题。

可以用f[i][j][st]表示推到第i行第j列时，行和列上rook的状态为st时方案的种数。

#include<stdio.h>
#include<string.h>
#define HASH 30007
#define SIZE 1000010
int N, K, rst, cst;
struct Hashmap
{
    int head[HASH], next[SIZE], state[SIZE], size;
    long long f[SIZE];
    void init()
    {
        memset(head, -1, sizeof(head));
        size = 0;
    }
    void push(int st, long long ans)
    {
        int i, h = st % HASH;
        for(i = head[h]; i != -1; i = next[i])
            if(st == state[i])
            {
                f[i] += ans;
                return ;
            }
        state[size] = st, f[size] = ans;
        next[size] = head[h];
        head[h] = size ++;
    }
}hm[2];
int encode(int rst, int cst)
{
    return (rst << N) | cst;
}
void decode(int st)
{
    cst = st & ((1 << N) - 1);
    rst = st >> N;
}
void dpblank(int i, int j, int cur)
{
    int k;
    for(k = 0; k < hm[cur].size; k ++)
    {
        decode(hm[cur].state[k]);
        hm[cur ^ 1].push(hm[cur].state[k], hm[cur].f[k]);
        if(((1 << i) & rst) == 0 && ((1 << j) & cst) == 0)
            hm[cur ^ 1].push(encode(rst | (1 << i), cst | (1 << j)), hm[cur].f[k]);
    }
}
void solve()
{
    int i, j, k, cur = 0;
    long long ans = 0;
    hm[cur].init();
    hm[cur].push(0, 1);
    for(i = 0; i < N; i ++)
        for(j = 0; j < N; j ++)
        {
            hm[cur ^ 1].init();
            dpblank(i, j, cur);
            cur ^= 1;
        }
    for(i = 0; i < hm[cur].size; i ++)
    {
        decode(hm[cur].state[i]);
        for(j = k = 0; j < N; j ++)
            if(rst & (1 << j))
                ++ k;
        if(k == K)
            ans += hm[cur].f[i];
    }
    printf("%I64d\n", ans);
}
int main()
{
    while(scanf("%d%d", &N, &K) == 2)
    {
        if(K > N)
            printf("0\n");
        else
            solve();
    }
    return 0;
}