hdu 1081 To The Max 【最大子矩阵和】

To The Max

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 8882 Accepted Submission(s): 4288

Problem 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
15

Source
Greater New York 2001

代码：

#include<stdio.h>  #include<iostream>  #include<math.h>  #include<stdlib.h>  #include<ctype.h>  #include<algorithm>  #include<vector>  #include<string.h>  #include<queue>  #include<stack>  #include<set>  #include<map>  #include<sstream>  #include<time.h>  #include<malloc.h>  using namespace std;  const int MAXN = 1010;int n;int p[MAXN][MAXN];int longmax(int a[],int n){    int b = 0;    int ans = -10000000;    for(int i=0;i<n;i++)    {        if (b>0)            b+= a[i];        else b= a[i];        ans = max (ans,b);    }    return ans;}int work(){    int t[1010];    int ans = -10000000;    for(int i=0;i<n;i++)    {        memset(t,0,sizeof(t));        for(int j = i;j<n;j++)//枚举第i到第j行的所有可能矩阵的和        {            int k;            for(k=0;k<n;k++)            {                t[k] += p[j][k];            }            int tt = longmax(t,k);            ans = max(ans,tt);        }    }    return ans;}int main(){    while(scanf("%d",&n)!=EOF)    {        for(int i=0;i<n;i++)            for(int j=0;j<n;j++)            {                scanf("%d",&p[i][j]);            }            int ans = work();            printf("%d\n",ans);    }    return 0;}