ZOJ 3213 Beautiful Meadow

ZOJ_3213

    我是用最小表示法写的这个插头dp，相比于回路问题来讲，变简单的地方是不用考虑合并两个相同的连通分量这一情况了，变复杂的地方就是多出了两个个独立插头。

    为了搞定独立插头的问题，我在编码的时候额外加了一个状态num，表示当前已经使用的独立插头的数目，这样只要保证num不超过2而且不合并相同的连通分量就能保证最后是一条简单路径。

    在dp的时候因为多了独立插头，所以分的情况就更多一些。

#include<stdio.h>
#include<string.h>
#define MAXD 15
#define HASH 30007
#define SIZE 1000010
int N, M, maze[MAXD][MAXD], code[MAXD], ch[MAXD], num, ans;
struct Hashmap
{
    int head[HASH], next[SIZE], state[SIZE], f[SIZE], size;
    void init()
    {
        memset(head, -1, sizeof(head));
        size = 0;
    }
    void push(int st, int ans)
    {
        int i, h = st % HASH;
        for(i = head[h]; i != -1; i = next[i])
            if(st == state[i])
            {
                if(ans > f[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, int st)
{
    int i;
    num = st & 7;
    for(i = m; i >= 0; i --)
    {
        st >>= 3;
        code[i] = st & 7;
    }
}
int encode(int *code, int m)
{
    int i, cnt = 1, 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 |= num;
    return st;
}
void init()
{
    int i, j;
    scanf("%d%d", &N, &M);
    ans = 0;
    memset(maze, 0, sizeof(maze));
    for(i = 1; i <= N; i ++)
        for(j = 1; j <= M; j ++)
        {
            scanf("%d", &maze[i][j]);
            if(maze[i][j] > ans)
                ans = maze[i][j];
        }
}
void dpblank(int i, int j, int cur)
{
    int k, left, up, t;
    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)
        {
            if(left != up)
            {
                code[j - 1] = code[j] = 0;
                for(t = 0; t <= M; t ++)
                    if(code[t] == up)
                        code[t] = left;
                hm[cur ^ 1].push(encode(code, j == M ? M - 1 : M), hm[cur].f[k] + maze[i][j]);
            }
        }
        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] + maze[i][j]);
            }
            if(maze[i + 1][j])
            {
                code[j - 1] = left + up, code[j] = 0;
                hm[cur ^ 1].push(encode(code, j == M ? M - 1 : M), hm[cur].f[k] + maze[i][j]);
            }
            if(num < 2)
            {
                ++ num, code[j - 1] = code[j] = 0;
                hm[cur ^ 1].push(encode(code, j == M ? M - 1 : M), hm[cur].f[k] + maze[i][j]);
            }
        }
        else
        {
            code[j - 1] = code[j] = 0;
            hm[cur ^ 1].push(encode(code, j == M ? M - 1 : M), hm[cur].f[k]);
            if(maze[i][j + 1] && maze[i + 1][j])
            {
                code[j - 1] = code[j] = 13;
                hm[cur ^ 1].push(encode(code, M), hm[cur].f[k] + maze[i][j]);
            }
            if(num < 2)
            {
                ++ num;
                if(maze[i + 1][j])
                {
                    code[j - 1] = 13, code[j] = 0;
                    hm[cur ^ 1].push(encode(code, j == M ? M - 1 : M), hm[cur].f[k] + maze[i][j]);
                }
                if(maze[i][j + 1])
                {
                    code[j - 1] = 0, code[j] = 13;
                    hm[cur ^ 1].push(encode(code, M), hm[cur].f[k] + maze[i][j]);
                }
            }
        }
    }
}
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;
        hm[cur ^ 1].push(encode(code, j == M ? M - 1 : M), hm[cur].f[k]);
    }
}
void solve()
{
    int i, j, cur = 0;
    hm[cur].init();
    hm[cur].push(0, 0);
    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 ++)
        if(hm[cur].f[i] > ans)
            ans = hm[cur].f[i];
    printf("%d\n", ans);
}
int main()
{
    int t;
    scanf("%d", &t);
    while(t --)
    {
        init();
        solve();
    }
    return 0;
}