FZU 1977 Pandora adventure

FZU_1977

    对于必须要走的点，那么在那个点就一定要有插头。而对于不必走的点，如果dp到该点时没有插头，那么既可以在这个点放两个插头开辟一个新的连通分量，也可以不放插头，相当于不经过这个点。

    此外由于回头只能有一条，而形成回路的最后一个点又是不确定的，所以可以再额外记录一下当前是否已经形成了回路，这样如果已经形成了回路而后面又遇到了插头或者必须要走的点，那么这个方案自然就是不合法的。

#include<stdio.h>

#include<string.h>

#define MAXD 15

#define HASH 30007

#define SIZE 500010

int N, M, maze[MAXD][MAXD], ch[MAXD], code[MAXD], isend;

char b[MAXD];

struct Hashmap

{

    int head[HASH], next[SIZE], size;

    long long state[SIZE], f[SIZE];

    void init()

    {

        memset(head, -1, sizeof(head));

        size = 0;

    }

    void push(long long 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];

void decode(int *code, int m, long long st)

{

    int i;

    isend = st & 7;

    for(i = m; i >= 0; i --)

    {

        st >>= 3;

        code[i] = st & 7;

    }

}

long long encode(int *code, int m)

{

    int i, cnt = 1;

    long long st = 0;

    memset(ch, -1, sizeof(ch));

    ch[0] = 0;

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

    {

        if(ch[code[i]] == -1)

            ch[code[i]] = cnt ++;

        code[i] = ch[code[i]];

        st <<= 3;

        st |= code[i];

    }

    st <<= 3;

    st |= isend;

}

void init()

{

    int i, j;

    scanf("%d%d", &N, &M);

    memset(maze, 0, sizeof(maze));

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

    {

        scanf("%s", b + 1);

        for(j = 1; j <= M; j ++)

        {

            if(b[j] == 'O')

                maze[i][j] = 2;

            else if(b[j] == '*')

                maze[i][j] = 1;

        }

    }

}

void shift(int *code, int m)

{

    int i;

    for(i = m; i > 0; i --)

        code[i] = code[i - 1];

    code[0] = 0;

}

void dpblock(int i, int j, int cur)

{

    int k;

    for(k = 0; k < hm[cur].size; k ++)

    {

        decode(code, M, hm[cur].state[k]);

        code[j - 1] = code[j] = 0;

        if(j == M)

            shift(code, M);

        hm[cur ^ 1].push(encode(code, M), hm[cur].f[k]);

    }

}

void dpblank(int i, int j, int cur)

{

    int k, t, left, up;

    for(k = 0; k < hm[cur].size; k ++)

    {

        decode(code, M, hm[cur].state[k]);

        left = code[j - 1], up = code[j];

        if(isend)

        {

            if(left || up || maze[i][j] == 2)

                continue;

            code[j - 1] = code[j] = 0;

            if(j == M)

                shift(code, M);

            hm[cur ^ 1].push(encode(code, M), hm[cur].f[k]);

            continue;

        }

        if(left && up)

        {

            if(left == up)

            {

                code[j - 1] = code[j] = 0, isend = 1;

                if(j == M)

                    shift(code, M);

                hm[cur ^ 1].push(encode(code, M), hm[cur].f[k]);

            }

            else

            {

                code[j - 1] = code[j] = 0;

                for(t = 0; t <= M; t ++)

                    if(code[t] == up)

                        code[t] = left;

                if(j == M)

                    shift(code, M);

                hm[cur ^ 1].push(encode(code, M), hm[cur].f[k]);

            }

        }

        else if(left || up)

        {

            if(maze[i][j + 1])

            {

                code[j - 1] = 0, code[j] = left + up;

                hm[cur ^ 1].push(encode(code, M), hm[cur].f[k]);

            }

            if(maze[i + 1][j])

            {

                code[j - 1] = left + up, code[j] = 0;

                if(j == M)

                    shift(code, M);

                hm[cur ^ 1].push(encode(code, M), hm[cur].f[k]);

            }

        }

        else

        {

            if(maze[i + 1][j] && maze[i][j + 1])

            {

                code[j - 1] = code[j] = 13;

                hm[cur ^ 1].push(encode(code, M), hm[cur].f[k]);

            }

            if(maze[i][j] == 1)

            {

                code[j - 1] = code[j] = 0;

                if(j == M)

                    shift(code, M);

                hm[cur ^ 1].push(encode(code, M), hm[cur].f[k]);

            }

        }

    }

}

void solve()

{

    int i, j, cur = 0;

    long long ans = 0;

    hm[cur].init();

    hm[cur].push(0, 1);

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

        for(j = 1; j <= M; j ++)

        {

            hm[cur ^ 1].init();

            if(maze[i][j] == 0)

                dpblock(i, j, cur);

            else

                dpblank(i, j, cur);

            cur ^= 1;

        }

    for(i = 0; i < hm[cur].size; i ++)

        ans += hm[cur].f[i];

    printf("%I64d\n", ans);

}

int main()

{

    int t, tt;

    scanf("%d", &t);

    for(tt = 0; tt < t; tt ++)

    {

        init();

        printf("Case %d: ", tt + 1);

        solve();

    }

    return 0;

}