HDU 1693 Eat the Trees

HDU_1639

    这个题目可以用插头dp的思想去做，由于允许多条回路，那么就只需要用0、1记录每个位置是否有插头，然后逐渐生成合法的状态并记录方案种数。

    更多和插头dp相关的题目可以参考胡浩的博客：http://www.notonlysuccess.com/index.php/plug-dp-complete/。

#include<stdio.h>
#include<string.h>
#define HASH 10007
#define STATE 1000010
#define MAXD 15
int N, M, code[MAXD], maze[MAXD][MAXD];
struct Hashmap
{
    int head[HASH], next[STATE], state[STATE], size;
    long long f[STATE];
    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 ;
            }
        f[size] = ans, state[size] = st;
        next[size] = head[h];
        head[h] = size ++;
    }
}hm[2];
void decode(int *code, int m, int st)
{
    int i;
    for(i = m; i >= 0; i --)
    {
        code[i] = st & 1;
        st >>= 1;
    }
}
int encode(int *code, int m)
{
    int i, st = 0;
    for(i = 0; i <= m; i ++)
    {
        st <<= 1;
        st |= code[i];
    }
    return st;
}
void init()
{
    int i, j, k;
    scanf("%d%d", &N, &M);
    memset(maze, 0, sizeof(maze));
    for(i = 1; i <= N; i ++)
        for(j = 1; j <= M; j ++)
            scanf("%d", &maze[i][j]);
}
void shift(int *code, int m)
{
    int i;
    for(i = m; i > 0; i --)
        code[i] = code[i - 1];
    code[0] = 0;
}
void dpblank(int i, int j, int cur)
{
    int k, 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(left && up)
        {
            code[j - 1] = code[j] = 0;
            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] = 1;
                hm[cur ^ 1].push(encode(code, M), hm[cur].f[k]);
            }
            if(maze[i + 1][j])
            {
                code[j - 1] = 1, code[j] = 0;
                if(j == M)
                    shift(code, M);
                hm[cur ^ 1].push(encode(code, M), hm[cur].f[k]);
            }
        }
        else
        {
            if(maze[i][j + 1] && maze[i + 1][j])
            {
                code[j - 1] = code[j] = 1;
                hm[cur ^ 1].push(encode(code, M), hm[cur].f[k]);
            }
        }
    }
}
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 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])
                dpblank(i, j, cur);
            else
                dpblock(i, j, cur);
            cur ^= 1;
        }
    for(i = 0; i < hm[cur].size; i ++)
        ans += hm[cur].f[i];
    printf("There are %I64d ways to eat the trees.\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;
}