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;

}