BZOJ#1084[SCOI2005]最大子矩阵

[SCOI2005]最大子矩阵

思路：

f[i][j][k]表示前i行中选择了j个子矩阵，并且第i行的状态式k的j个最大子矩阵的和；
k有四种方式，10，01，11，11（两个不属于一个矩阵）
要么承接上一层，要么单独为一个矩阵

代码：

#include <bits/stdc++.h>
#define int long long
int _ = 0, Case = 1;
using namespace std;
#define all(v) begin(v),end(v)
#define nline '\n'
const int N = 110;
int a[N][3];
int f[N][20][5];

void solve(int Case) {
    int n, m, k;
    cin >> n >> m >> k;
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= m; j++) {
            cin >> a[i][j];
        }
    }
    memset(f, 0xcf, sizeof f);
    if (m == 1) {
        f[0][0][0] = 0;
        //f[0][0][1] = 0;
        for (int i = 1; i <= n; i++) {
            for (int j = 0; j <= k; j++) {
                f[i][j][0] = max(f[i - 1][j][1], f[i - 1][j][0]);
                f[i][j][1]=f[i-1][j][1]+a[i][1];
                if (j)
                    f[i][j][1] = max(max(f[i - 1][j - 1][0], f[i - 1][j - 1][1])+a[i][1],f[i][j][1]);
            }
        }
        cout << max(f[n][k][1], f[n][k][0]) << nline;
    }
    else {
        f[0][0][0]=0;
        for (int i = 1; i <= n; i++) {
            for (int j = 0; j <= k; j++) {
                for (int t = 0; t <= 4; t++) {
                    f[i][j][0] = max(f[i][j][0], f[i - 1][j][t]);
                }
                f[i][j][1] = max(f[i - 1][j][1], f[i - 1][j][4]) + a[i][1];
                for (int t = 0; t <= 4; t++) {
                    if (j)
                        f[i][j][1] = max(f[i - 1][j - 1][t] + a[i][1], f[i][j][1]);
                }
                f[i][j][2] = max(f[i - 1][j][2], f[i - 1][j][4]) + a[i][2];
                for (int t = 0; t <= 4; t++) {
                    if (j)
                        f[i][j][2] = max(f[i - 1][j - 1][t] + a[i][2], f[i][j][2]);
                }
                f[i][j][3] = max(f[i - 1][j][3] + a[i][1] + a[i][2], f[i][j][3]);
                for (int t = 0; t <= 4; t++) {
                    if (j)
                        f[i][j][3] = max(f[i - 1][j - 1][t] + a[i][2] + a[i][1], f[i][j][3]);
                }
                f[i][j][4] = max(f[i - 1][j][4] + a[i][1] + a[i][2], f[i][j][4]);
                if (j)
                    f[i][j][4] = max(max(f[i - 1][j - 1][1], f[i - 1][j - 1][2]) + a[i][1] + a[i][2], f[i][j][4]);
                for (int t = 0; t <= 4; t++) {
                    if (j >= 2)
                        f[i][j][4] = max(f[i - 1][j - 2][t] + a[i][2] + a[i][1], f[i][j][4]);
                }
            }
        }
        int res = 0;
        for (int i = 0; i <= 4; i++) res = max(f[n][k][i], res);
        cout << res << nline;
    }



}

signed main() {
    ios::sync_with_stdio(false); cin.tie(nullptr);
//   cin >> _; for (Case = 1; Case <= _; Case++)
    solve(Case);

    return 0;
}