ZOJ 1074 Problem Set || nyoj 104 最大和

To the Max

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Time Limit: 2 Seconds      Memory Limit: 65536 KB

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Problem

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.

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.

Example

Input

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

Output

15

 求最大子矩阵和, 可以用求最大和连续子序列解，将矩阵二维转化成一维， 压缩行或列都可以，毕竟n*n的都一样。

ps:一维数组最大连续子序列和

        int a[n],b=0,sum=-0x3f3f3f3f;        for(int i=0;i<n;i++)        {             if(b>0) b+=a[i];             else b=a[i];             if(b>sum) sum=b;          }        printf("%d\n",sum);

 

 循环DP，共三层循环。
最外层i循环1-N，表明子矩阵是从第i列开始累加的。
第二层j循环i-N，表明子矩阵是从第i列累加到第j列。
第三层k从1到M做一维DP

#include<cstdio>#include<iostream>#include<cstring>#include<map>#include<algorithm>using namespace std;int main(){    int n,a[110][110],max,sum,i,j,k;    while(scanf("%d",&n)!=EOF)    {        for(i = 1 ; i <= n ; i++)        {            for( j = 1 ; j <= n ; j++)            {                int t;                scanf("%d",&t);                a[i][j]=a[i][j-1]+t;                //表示第i行从第1列到第j列之和            }        }        max = 0 ;        for(i = 1; i <= n ; i++)//i表示列        {            for(j = i ; j <= n ; j++)//j表示列，中间变量            {                        // j > i                sum  =  0;                for(k = 1 ; k <= n ; k++)//表示行，从1到第n行                  {                    int t;                    t = a[k][j]-a[k][i-1];//表示从第k行i列到k行j列之和,                                            //j-(i-1)表示区间i--j                    sum +=t;               //状态压缩至一维求最大和                    if(sum < 0 ) sum =  0;                    if(sum > max) max = sum ;                }//k每循环完一次，sum表示第i到j列在第1到n行中的最大连续子序列和            }        }        printf("%d\n",max);    }    return 0;}