多唯DP

Code

#pragma GCC target("avx,sse2,sse3,sse4,popcnt")
#pragma GCC optimize("O2,O3,Ofast,inline,unroll-all-loops,-ffast-math")
#include <bits/stdc++.h>
using namespace std;
#define js ios::sync_with_stdio(false);cin.tie(0); cout.tie(0)
#define all(vv) vv.begin(), vv.end()
#define endl "\n"
typedef long long ll; typedef unsigned long long ull; typedef long double ld;
const ll MOD = 1e9 + 7;
inline ll read() { ll s = 0, w = 1; char ch = getchar(); for (; !isdigit(ch); ch = getchar()) if (ch == '-') w = -1; for (; isdigit(ch); ch = getchar())    s = (s << 1) + (s << 3) + (ch ^ 48); return s * w; }
inline void write(ll x) { if (!x) { putchar('0'); return; } char F[40]; ll tmp = x > 0 ? x : -x; if (x < 0)putchar('-');    int cnt = 0;    while (tmp > 0) { F[cnt++] = tmp % 10 + '0';        tmp /= 10; }    while (cnt > 0)putchar(F[--cnt]); }
inline ll gcd(ll x, ll y) { return y ? gcd(y, x % y) : x; }
ll qpow(ll a, ll b) { ll ans = 1;    while (b) { if (b & 1)    ans *= a;        b >>= 1;        a *= a; }    return ans; }    ll qpow(ll a, ll b, ll mod) { ll ans = 1; while (b) { if (b & 1)(ans *= a) %= mod; b >>= 1; (a *= a) %= mod; }return ans % mod; }
inline int lowbit(int x) { return x & (-x); }

const int N = 100 + 7;
int s[N][3], dp[N][N][11];

int main() {
    int n = read(), m = read(), k = read();
    for (int i = 1; i <= n; ++i)
        for (int j = 1; j <= m; ++j)
            s[i][j] = s[i - 1][j] + read();
    for (int i = 1; i <= n; ++i) // 第一列的末尾选取
        for (int j = 1; j <= n; ++j) //第二列
            for (int a = 1; a <= k; ++a) { //矩阵个数
                dp[i][j][a] = max(dp[i - 1][j][a], dp[i][j - 1][a]);
                for (int b = 0; b < i; ++b)    //第一列状态转移
                    dp[i][j][a] = max(dp[i][j][a], dp[b][j][a - 1] + s[i][1] - s[b][1]);
                for (int b = 0; b < i; ++b)    //第二列状态转移
                    dp[i][j][a] = max(dp[i][j][a], dp[i][b][a - 1] + s[j][2] - s[b][2]);
                if (i == j)
                    for (int b = 0; b < i; ++b) //选底长为2的矩阵转移
                        dp[i][j][a] = max(dp[i][j][a], dp[b][b][a - 1] + s[i][1] + s[j][2] - s[b][1] - s[b][2]);
            }
    write(dp[n][n][k]);
    return 0;
}