AcWing 276. I-区域

题目链接

设 (0) 为单调伸长， (1) 为单调伸短。

设 (f[i][j][l][r][x(0 / 1)][y (0 / 1)]) 为第 (i) 行，已经选出(j)个格子，第(i)行选择了([l, r]) 区间的最大值。左右端点(x, y)分别为 单调伸长 / 单调伸短 的最大权值。

状态转移：

1. 若 (x = 0, y = 0)，则 (f[i][j][l][r][x][y] = max{f[i - 1][j - (r - l + 1)][l'][r'][0][0]} + cost(i, l, r)) ，其中(l <= l' <= r' <= r, j >= r - l + 1)。

2. 若 (x = 0, y = 1), 则 (f[i][j][l][r][x][y] = max{f[i - 1][j - (r - l + 1)][l'][r'][0][0 / 1]} + cost(i, l, r))，其中(l <= l' <= r <= r', j >= r - l + 1)。

3. 若 (x = 1, y = 0), 则 (f[i][j][l][r][x][y] = max{f[i - 1][j - (r - l + 1)][l'][r'][0 / 1][0]} + cost(i, l, r))，其中(l' <= l <= r' <= r, j >= r - l + 1) 。

4. 若 (x = 1, y = 1), 则 (f[i][j][l][r][x][y] = max{f[i - 1][j - (r - l + 1)][l'][r'][0 / 1][0 / 1]} + cost(i, l, r))，其中(l' <= l <= r <= r', j >= r - l + 1) 。

初始状态： (f[i][0][l][r][0][0] = 0 (0 <= i <= n, 1 <= l <= r <= m))，其余为负无穷。

目标：(max{f[i][k][l][r][0 / 1][0 / 1]} (1 <= i <= n))

输出方案只需通过状态转移延展即可。(cost(i, l, r)) 用前缀和计算​

时间复杂度(O(n ^ 7))

C++ 代码

#include <cstdio>
#include <iostream>
#include <cstring>
using namespace std;
const int N = 16;

struct S{
    int i, j, l, r, x, y;
}pre[N][N * N][N][N][2][2], t;

int n, m, k, a[N][N], f[N][N * N][N][N][2][2];
int ans = 0;

int inline cost(int i, int l, int r){
    return a[i][r] - a[i][l - 1];
}

void print(S x){
    if(x.j == 0) return;
    print(pre[x.i][x.j][x.l][x.r][x.x][x.y]);
    for(int i = x.l; i <= x.r; i++) 
        printf("%d %d
", x.i, i);
}

int main(){
    memset(f, 0xcf, sizeof f);
    
    scanf("%d%d%d", &n, &m, &k);
    for(int r = 0; r <= n; r++) {
        for(int i = 1; i <= m; i++){
            for(int j = i; j <= m; j++)
                f[r][0][i][j][0][0] = 0;
        }
    }
    for(int i = 1; i <= n; i++)
        for(int j = 1; j <= m; j++)
            scanf("%d", &a[i][j]), a[i][j] += a[i][j - 1];

    for(int i = 1; i <= n; i++){
        for(int j = 1; j <= k; j++){
            for(int l = 1; l <= m; l++){
                for(int r = l; r <= m; r++){
                    if(j < r - l + 1) continue;
                    
                    //x = 0, y = 0;
                    for(int l1 = l; l1 <= r; l1++){
                        for(int r1 = l1; r1 <= r; r1++){
                            int &v = f[i][j][l][r][0][0], val = f[i - 1][j - (r - l + 1)][l1][r1][0][0] + cost(i, l, r);
                            if(v < val) {
                                v = val, pre[i][j][l][r][0][0] = (S){i - 1, j - (r - l + 1), l1, r1, 0, 0};
                            }
                        }
                    }
                    //x = 0, y = 1;
                    for(int l1 = l; l1 <= r; l1++){
                        for(int r1 = r; r1 <= m; r1++){
                            for(int y1 = 0; y1 < 2; y1++) {
                                int &v = f[i][j][l][r][0][1], val = f[i - 1][j - (r - l + 1)][l1][r1][0][y1] + cost(i, l, r);
                                if(v < val) {
                                    v = val, pre[i][j][l][r][0][1] = (S){i - 1, j - (r - l + 1), l1, r1, 0, y1};
                                }
                            }
                        }
                    }

                    // x = 1, y = 0;
                    for(int l1 = 1; l1 <= l; l1++){
                        for(int r1 = l; r1 <= r; r1++){
                            for(int x1 = 0; x1 < 2; x1++) {
                                int &v = f[i][j][l][r][1][0], val = f[i - 1][j - (r - l + 1)][l1][r1][x1][0] + cost(i, l, r);
                                if(v < val) {
                                    v = val, pre[i][j][l][r][1][0] = (S){i - 1, j - (r - l + 1), l1, r1, x1, 0};
                                }   
                            }
                        }
                    }

                    // x = 1, y = 1;
                    for(int l1 = 1; l1 <= l; l1++){
                        for(int r1 = r; r1 <= m; r1++){
                            for(int x1 = 0; x1 < 2; x1++) {
                                for(int y1 = 0; y1 < 2; y1++) {
                                    int &v = f[i][j][l][r][1][1], val =  f[i - 1][j - (r - l + 1)][l1][r1][x1][y1] + cost(i, l, r);
                                    if(v < val) {
                                        v = val, pre[i][j][l][r][1][1] = (S){i - 1, j - (r - l + 1), l1, r1, x1, y1};
                                    }
                                }
                            }
                        }
                    }
                    if(j == k){
                        for(int x = 0; x < 2; x++) {
                            for(int y = 0; y < 2; y++) {
                                if(ans < f[i][j][l][r][x][y]) {
                                    ans = f[i][j][l][r][x][y], t = (S){i, j, l, r, x, y};
                                }
                            }
                        }
                    }
                }
            }
        }
    }
    printf("Oil : %d
", ans);
    print(t);
    return 0;   
}