To the Max（DP）

最新推荐文章于 2024-07-31 10:00:00 发布

iteye_6881

最新推荐文章于 2024-07-31 10:00:00 发布

阅读量139

点赞数

CC 4.0 BY-SA版权

分类专栏： POJ

本文链接：https://blog.youkuaiyun.com/iteye_6881/article/details/82602891

POJ 专栏收录该内容

102 篇文章

订阅专栏

本文介绍了一种通过动态规划解决最大子矩阵和问题的方法，包括输入格式、输出要求及详细算法步骤。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

To the Max

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 40625		Accepted: 21525

Description

Given a two-dimensional array of positive and negative integers, a sub-rectangle is any contiguous sub-array of size 1*1 or greater located within the whole array. The sum of a rectangle is the sum of all the elements in that rectangle. In this problem the sub-rectangle with the largest sum is referred to as the maximal sub-rectangle.
As an example, the maximal sub-rectangle of the array:

0 -2 -7 0
9 2 -6 2
-4 1 -4 1
-1 8 0 -2
is in the lower left corner:

9 2
-4 1
-1 8
and has a sum of 15.

Input

The input consists of an N * N array of integers. The input begins with a single positive integer N on a line by itself, indicating the size of the square two-dimensional array. This is followed by N^2 integers separated by whitespace (spaces and newlines). These are the N^2 integers of the array, presented in row-major order. That is, all numbers in the first row, left to right, then all numbers in the second row, left to right, etc. N may be as large as 100. The numbers in the array will be in the range [-127,127].

Output

Output the sum of the maximal sub-rectangle.

Sample Input

Sample Output

题意：

给出 N（1 ~ 100），代表有一个 N X N 的矩阵，后给出这个 N X N 的矩阵，输出这个矩阵的最大子矩阵和。

思路：

DP。最大子段和是 dp [ i ] = max { dp [ i - 1 ] + num [ i ], num[ i ] }，dp [ i ] 代表到 i 为止最大的子段和。

对于矩阵的话，枚举行的首尾位置，后把列浓缩一个总和，走一遍最大字段和，同时比较出最大值即可。

AC：

#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

const int INF = -100000000;
int num[105][105], sum[105][105];
int dp[105], ans[105];
int n;

int solve() {
    memset(dp, 0, sizeof(dp));

    int Max = INF;
    for (int i = 1; i <= n; ++i) {
        dp[i] = max(ans[i], dp[i - 1] + ans[i]);
        Max = max(Max, dp[i]);
    }

    return Max;
}

int main() {

    scanf("%d", &n);
    memset(sum, 0, sizeof(sum));
    for (int i = 1; i <= n; ++i) {
        for (int j = 1; j <= n; ++j) {
            scanf("%d", &num[i][j]);
            sum[i][j] = sum[i - 1][j] + num[i][j];
        }
    }

    int Max = INF;
    for (int i = 1; i <= n; ++i) {
        for (int j = i; j <= n; ++j) {

            for (int k = 1; k <= n; ++k) {
                ans[k] = sum[j][k] - sum[i - 1][k];
            }

            Max = max(Max, solve());
        }
    }

    printf("%d\n", Max);

    return 0;
}