POJ 1050 To the Max

To the Max Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 48349 Accepted: 25582

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

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

Sample Output

15

Source

Greater New York 2001 ---------------------------------华丽的分割线------------------------------- 问题描述： 给定一个N阶方阵（正负不定），求其n阶子阵和的最大值（子阵必须是连在一起的）。 解体思路： DP, 求最大字段和。题目给出是二维的，故降低其维数以一维字段和的方法解决。 Code： #include<stdio.h> #include<cstring> #define MAX 101 int main() { int i,j,k,n,m,ans,max1,sum1,line[MAX][MAX]; scanf("%d",&n); memset(line,0,sizeof(line)); for(i=1;i<=n;i++) for(j=1;j<=n;j++){ scanf("%d",&m); line[i][j]=line[i][j-1]+m;//line[i][j]为第i行前j个元素的和 } max1=0; for(i=0;i<=n;i++) for(j=i,sum1=0;j<=n;sum1=0,j++)//这里是对列区间[i+1,j]的枚举，故此降低了维数 for(k=1;k<=n;k++){ if(sum1<0)sum1=0; sum1+=line[k][j]-line[k][i];//以这个差为新的数列（区间内的和），对行数求最大字段和 if(sum1>max1)max1=sum1; } printf("%d\n",max1); return 0; }