HDU_1964

    这个题目只需要把求回路数量的dp方程改写成取最优解的dp方程即可。

    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 MAXD 15
#define HASH 30007
#define SIZE 1000010
#define INF 0x3f3f3f3f
int N, M, code[MAXD], ch[MAXD], maze[MAXD][MAXD], rcost[MAXD][MAXD], dcost[MAXD][MAXD];
char b[50];
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(state[i] == st)
            {
                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;
    for(i = m; i >= 0; i --)
    {
        code[i] = st & 7;
        st >>= 3;
    }
}
int encode(int *code, int m)
{
    int i, cnt = 0, 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];
    }
    return st;
}
void init()
{
    int i, j, k;
    gets(b), sscanf(b, "%d%d", &N, &M);
    memset(maze, 0, sizeof(maze));
    for(i = 1; i <= N; i ++)
        for(j = 1; j <= M; j ++)
            maze[i][j] = 1;
    gets(b);
    for(i = 1; i < 2 * N; i ++)
    {
        gets(b);
        if(i & 1)
        {
            for(j = 1, k = 2; k < 2 * M; j ++, k += 2)
                rcost[(i + 1) >> 1][j] = b[k] - '0';
        }
        else
        {
            for(j = 1, k = 1; k < 2 * M; j ++, k += 2)
                dcost[i >> 1][j] = b[k] - '0';
        }
    }
    gets(b);
}
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, 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)
            {
                if(i == N && j == M)
                {
                    code[j - 1] = code[j] = 0;
                    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] + rcost[i][j]);
            }
            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] + dcost[i][j]);
            }
        }
        else
        {
            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] + rcost[i][j] + dcost[i][j]);
            }
        }
    }
}
void solve()
{
    int i, j, cur = 0, ans = INF;
    hm[cur].init();
    hm[cur].push(0, 0);
    for(i = 1; i <= N; i ++)
        for(j = 1; j <= M; j ++)
        {
            hm[cur ^ 1].init();
            dpblank(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;
    gets(b), sscanf(b, "%d", &t);
    while(t --)
    {
        init();
        solve();
    }
    return 0;
}
posted on 2012-04-20 00:59  Staginner  阅读(318)  评论(0编辑  收藏  举报