HDU3377 Plan 插头dp

HDU3377 Plan 插头dp

原题：HDU3377

题意：$N*M$矩阵，格子权值有正负，求从左上角到右下角的权值和最大的路径。

解：简单路径问题插头dp。
在起点终点位置维护一个独立插头
对起点，没有左，上插头的状态转移到右独立插头或者下独立插头。
对终点，只有左独立插头或只有右独立插头的状态转移到没有右，下插头。
对普通格子，正常处理，有左，上插头合并连通分量，只有左插头或上插头延续连通分量，没有左，上插头，新建连通分量或者不选此格。
这题不做处理，使用最小表示法可能会T，所以可以旋转矩阵，使得m为$min(n, m)$。

代码如下：

#include <cstdio>
#include <iostream>
#include <cstring>
#include <queue>
#include <algorithm>
#include <cmath>
using namespace std;
#define N 15
#define HASH 10007
#define STATE 1000100
#define inf 0x7f7f7f7f
#define mod 1000000007
typedef long long ll;
int n, m;
int ch[N], code[N];
int mp[N][N], mp2[N][N];

struct HASHMAP
{
    int head[HASH], next[STATE], size;
    ll state[STATE], f[STATE];
    void init() {
        size = 0;
        memset(head, -1, sizeof(head));
    }
    void push(ll st, ll ans) {
        int h = st % HASH;
        for (int i = head[h]; i != -1; i = next[i]) {
            if (state[i] == st) {
                f[i] = max(ans, f[i]);
                return;
            }
        }
        state[size] = st;
        f[size] = ans;
        next[size] = head[h];
        head[h] = size++;
    }
}hm[2];

void decode(ll st) 
{
    for (int i = m; i >= 0; i--) {
        code[i] = st & 7;
        st >>= 3;
    }
}

ll encode()
{
    int cnt = 1;
    ll st = 0;
    memset(ch, -1, sizeof(ch));
    ch[0] = 0;
    ch[7] = 7;
    for (int 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 shift()
{
    for (int i = m; i > 0; i--) {
        code[i] = code[i-1];
    }
    code[0] = 0;
}

void dp(int r, int c, int cur)
{
    for (int k = 0; k < hm[cur].size; k++) {
        decode(hm[cur].state[k]);
        int up = code[c], left= code[c-1];
        if ((r == 1 && c == 1) || (r == n && c == m)) {
            if (!up && !left) {
                if (c < m) {
                    code[c] = 7;
                    code[c-1] = 0;
                    hm[cur^1].push(encode(), hm[cur].f[k]+mp[r][c]);
                }
                if (r < n) {
                    code[c] = 0;
                    code[c-1] = 7;
                if (m == c) shift();
                    hm[cur^1].push(encode(), hm[cur].f[k]+mp[r][c]);
                }
            }
            if (up || left) {
                code[c] = code[c-1] = 0;
                if (m == c) shift();
                hm[cur^1].push(encode(), hm[cur].f[k]+mp[r][c]);
            }
            continue;
        }
        if (up && left) {
            if (up != left) {
                int t;
                t = max(up, left);
                code[c] = code[c-1] = 0;
                for (int i = 0; i <= m; i++) {
                    if (code[i] == left || code[i] == up) {     //注意不是code[i] == up，wa了无数发
                        code[i] = t;
                    }
                }
                if (m == c) shift();
                hm[cur^1].push(encode(), hm[cur].f[k]+mp[r][c]);
            }
        }
        else if (up || left) {
            if (c < m) {
                code[c-1] = 0;
                code[c] = up | left;
                if (m == c) shift();
                hm[cur^1].push(encode(), hm[cur].f[k]+mp[r][c]);
            }
            if (r < n) {
                code[c-1] = up | left;
                code[c] = 0;
                if (m == c) shift();
                hm[cur^1].push(encode(), hm[cur].f[k]+mp[r][c]); 
            }
        }
        else {
            if (c < m && r < n) {
                code[c-1] = code[c] = 6;
                // if (c == m) shift();
                hm[cur^1].push(encode(), hm[cur].f[k]+mp[r][c]);
            }
            code[c] = code[c-1] = 0;
            if (c == m) shift();
            hm[cur^1].push(encode(), hm[cur].f[k]);
        }
    }
}

int main()
{
    int cas = 0;
    while (~scanf("%d%d", &n, &m)) {
        for (int i = 1; i <= n; i++) {
            for (int j = 1; j <= m; j++) {
                scanf("%d", &mp[i][j]);
            }
        }
        if (n == 1 && m == 1) {
            printf("Case %d: %d\n", ++cas, mp[1][1]);
            continue;
        }
        if (n < m) {
            for (int i = 1; i <= m; i++) {
                for (int j = 1; j <= n; j++) {
                    mp2[i][j] = mp[j][i];
                }
            }
            for (int i = 1; i <= m; i++) {
                for (int j = 1; j <= n; j++) {
                    mp[i][j] = mp2[i][j];
                }
            }
            swap(n, m);
        }
        int cur = 0;
        hm[cur].init();
        hm[cur].push(0, 0);
        for (int i = 1; i <= n; i++) {
            for (int j = 1; j <= m; j++) {
                hm[cur^1].init();
                dp(i, j, cur);
                cur ^= 1;
            }
        }
        ll ans = -inf;
        for (int i = 0; i < hm[cur].size; i++) {
            ans = max(ans, hm[cur].f[i]);
        }
        printf("Case %d: %lld\n", ++cas, ans);
    }   
    return 0;
}