NOI1768: the largest sub-matrix

topic

1768: The biggest sub-matrix

answer

Violence to solve

Range of topics $0 < N <= 100$ , violence can be directly solved the biggest sub-matrix. Complexity $O (n ^ 4)$, this complexity is generally acceptable.

#include<stdio.h>
#include<string.h>
#include<stdlib.h>

#define maxn 105
#define inf 0x3f3f3f3f

int n, ans;
int f[maxn][maxn], a[maxn][maxn];

int _max(int a, int b){
    return a > b ? a : b;
}
int main()
{
    scanf("%d", &n);
    ans = -inf;
    int i, j, k, l;
    for(i = 1; i <= n; i++){
        for(j = 1; j <= n; j++){
            scanf("%d", &a[i][j]);
            f[i][j] = f[i - 1][j] + f[i][j - 1] - f[i - 1][j - 1] + a[i][j];
        }
    }
    for(i = 1; i <= n; i++){
        for(j = 1; j <= n; j++){
            for(k = i; k <= n; k++){
                for(l = j; l <= n; l++){
                    ans = _max(ans, f[k][l] - f[k][j - 1] - f[i - 1][l] + f[i - 1][j - 1]);
                }
            }
        }
    }
    printf("%d\n", ans);
    return 0;
}

dp solving

• $pre[i][j]$ means that for the first $j$ column before $i$ matrix of rows and prefix. which is $pre[i][j] = \sum_{k = 1}^{i}a[k][j]$。
• $f[k]$ represents the starting line number $i$ , end line numbers $j$ sub-matrices, the first sub-matrix $j$ and elements of the column.

Using a similar method to solve the maximum continuous substring $f[k]$ is solved. The resulting sub-matrix is the maximum, the complexity of $O (n ^ 3)$。

#include<stdio.h>
#include<string.h>
#include<stdlib.h>

#define maxn 105
#define inf 0x3f3f3f3f

int n, ans;
int pre[maxn][maxn], a[maxn][maxn], f[maxn];

int _max(int a, int b){
    return a > b ? a : b;
}
int main()
{
    scanf("%d", &n);
    ans = -inf;
    int i, j, k;
    for(i = 1; i <= n; i++){
        for(j = 1; j <= n; j++){
            scanf("%d", &a[i][j]);
            pre[i][j] = pre[i - 1][j] + a[i][j];
        }
    }
    for(i = 1; i <= n; i++){
        for(j = i + 1; j <= n; j++){
            for(k = 1; k <= n; k++){
                f[k] = pre[j][k] - pre[i - 1][k];
            }
            for(k = 1; k <= n; k++){
                f[k] = _max(f[k - 1] + f[k], f[k]);
                ans = _max(ans, f[k]);
            }
        }
    }
    printf("%d\n", ans);
    return 0;
}