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  • HDU 1964 Pipes

    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;
    }
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  • 原文地址:https://www.cnblogs.com/staginner/p/2458695.html
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