1084

思路：

dp[i][j][k]:第一列選前i個第二列選前j個總共選了k個子矩陣的最大值

注意空矩陣也算子矩陣

代碼:

#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); //headconst 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\n", ans);return 0; }

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轉載于:https://www.cnblogs.com/widsom/p/10295186.html

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