BZOJ 1084: [SCOI2005]最大子矩阵

1084

思路：

dp[i][j][k]:第一列选前i个第二列选前j个总共选了k个子矩阵的最大值

注意空矩阵也算子矩阵

代码:

#pragma GCC optimize(2)
#pragma GCC optimize(3)
#pragma GCC optimize(4)
#include<bits/stdc++.h>
using namespace std;
#define fi first
#define se second
#define pi acos(-1.0)
#define LL long long
//#define mp make_pair
#define pb push_back
#define ls rt<<1, l, m
#define rs rt<<1|1, m+1, r
#define ULL unsigned LL
#define pll pair<LL, LL>
#define pli pair<LL, int>
#define pii pair<int, int>
#define piii pair<pii, int>
#define pdd pair<long double, long double>
#define mem(a, b) memset(a, b, sizeof(a))
#define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define fopen freopen("in.txt", "r", stdin);freopen("out.txt", "w", stout);
//head

const int N = 105;
int dp[N][N][12], f[N][12];
int a[N][3], sum[N][3];
int main() {
    int n, m, k;
    scanf("%d %d %d", &n, &m, &k);
    for (int i = 1; i <= n; i++) for (int j = 1; j <= m; j++) scanf("%d", &a[i][j]);
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= m; j++)
            sum[i][j] = sum[i-1][j] + a[i][j];
    }
    int ans = 0;
    if(m == 1) {
        for (int i = 1; i <= n; i++) {
            for (int j = 1; j <= k; j++) {
                f[i][j] = f[i-1][j];
                for (int l = 0; l < i; l++)
                    f[i][j] = max(f[i][j], f[l][j-1] + sum[i][1] - sum[l][1]);
                ans = max(ans, f[i][j]);
            }
        }
    }
    else {
        for (int i = 1; i <= n; i++) {
            for (int j = 1; j <= n; j++) {
                for (int l = 1; l <= k; l++) {
                    dp[i][j][l] = max(dp[i-1][j][l], dp[i][j-1][l]);
                    for (int m = 0; m < i; m++) dp[i][j][l] = max(dp[i][j][l], dp[m][j][l-1] + sum[i][1] - sum[m][1]);
                    for (int m = 0; m < j; m++) dp[i][j][l] = max(dp[i][j][l], dp[i][m][l-1] + sum[j][2] - sum[m][2]);
                    if(i == j) for (int m = 0; m < i; m++) dp[i][j][l] = max(dp[i][j][l], dp[m][m][l-1] + sum[i][1] - sum[m][1] + sum[j][2] - sum[m][2]);
                    ans = max(ans, dp[i][j][l]);
                }
            }
        }
    }
    printf("%d
", ans);
    return 0;
}