POJ1050-To the Max

To the Max

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 48010 Accepted: 25395

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

40 -2 -7 0 9 2 -6 2-4 1 -4  1 -18  0 -2

Sample Output

15

Source

Greater New York 2001

题意：在一个方阵中求最大的子矩阵

解题思路：

先复习一下最大子序列和：用dp[i]表示以第i个数为结尾的子序列和最大值，状态转移方程：
dp[i]=a[i] (dp[i-1]<0)
dp[i]=dp[i-1]+a[i] (dp[i-1]>=0)
求最大子矩阵和的时候,就是取出两行i,j,把这两行之间同一列的都加起来形成另外一个数组,求这个数组的最大子段和,求出来的这个和,就是这两行之间高度为i-j的子矩阵中最大的和。

#include <iostream>#include <cstdio>#include <string>#include <cstring>#include <algorithm>#include <cmath>#include <queue>#include <vector>#include <set>#include <stack>#include <map>#include <climits>using namespace std;#define LL long longint n,a[105][105],x[105];int solve(){    int ma=-200,sum=0;    for(int i=1;i<=n;i++)    {        if(sum>=0) sum+=x[i];        else sum=x[i];        ma=max(ma,sum);    }    return ma;}int main(){    while(~scanf("%d",&n))    {        int ma=-200;        for(int i=1;i<=n;i++)        {            for(int j=1;j<=n;j++)                scanf("%d",&a[i][j]);        }        for(int i=1;i<=n;i++)        {            memset(x,0,sizeof x);            for(int j=i;j<=n;j++)            {                for(int k=1;k<=n;k++)                    x[k]+=a[j][k];                int ans=solve();                if(ma<ans) ma=ans;            }        }        printf("%d\n",ma);    }    return 0;}