HDU 1081 To The Max【DP】【最大子段矩阵求和】

最新推荐文章于 2020-09-06 16:19:06 发布

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最新推荐文章于 2020-09-06 16:19:06 发布

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分类专栏： ACM__DP ACM__最大子段矩阵求和 HDU__ 文章标签： ACM HDU dp

本文链接：https://blog.youkuaiyun.com/qq_24653023/article/details/50740622

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本文详细解析二维数组最大子矩形和问题，包括输入格式、输出要求及解决方法，通过实例演示如何利用动态规划求解该问题。

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D - To The Max

Crawling in process... Crawling failed Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u

Submit Status Practice HDU 1081

Description

Given a two-dimensional array of positive and negative integers, a sub-rectangle is any contiguous sub-array of size 1 x 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 x 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

     
      4
0 -2 -7 0 9 2 -6 2
-4 1 -4 1 -1
8 0 -2

Sample Output

注意：

1：由一维数组最大子段和求解方法引申到二位矩阵最大

一维数组最大子段和最大的方法是 temp[i]=(temp[i-1]>0?temp[i-1]:0)+a[i];

二维压缩成一维数组后再找出最大组合

/*
一道DP题目
二维数组压缩
*/
#include<stdio.h>
#include<string.h>
int max(int n,int a[])//一维数组最大子段和求解方法
{
    int temp[133];
    int max=n*(-127);
    memset(temp,0,sizeof(temp));
    for(int i=1; i<=n; i++)
    {
        temp[i]=(temp[i-1]>0?temp[i-1]:0)+a[i];
        if(max<temp[i])
        {
            max=temp[i];
        }
    }
    return max;
}
int main (void)
{
    int sum, n,a[130][130],b[130];
    while(~scanf("%d",&n)&&n)
    {
        int sum=n*(-127);
        memset(a,0,sizeof(a));
        for(int i=1; i<=n; i++)
        {
            for(int ii=1; ii<=n; ii++)
            {
                scanf("%d",&a[i][ii]);//输入
            }
        }
        for(int i=1; i<=n; i++)//从第一行开始
        {
            for(int ii=1; ii<=n; ii++)//每次更新i. b[]刷新
            {
                b[ii]=0;
            }
            for(int j=i; j<=n; j++)
            {
                for(int jj=1; jj<=n; jj++)
                {
                    b[jj]+=a[j][jj];//表示从第i行到第n行中第k列的总和。
                    //这两句放在这里有问题 WA
                    //int maxs=max(n,b);
                    //if(maxs>sum) sum=maxs;
                }
                int maxs=max(n,b);
                if(maxs>sum) sum=maxs;
            }
        }
        printf("%d\n",sum);
    }
    return 0;
}