hdu 1081 To The Max最大矩阵和(动态规划)

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

#include<stdio.h>int f(int n,int *a){    int i;    int b=0,sum=0;    for (i=1;i<=n;i++)    {        if (b>0)            b+=a[i];        else            b=a[i];        if (b>sum)            sum=b;    }    return sum;}int sum (int n,int a[][150]){    int sum=0;     int i,j,k;     int b[150];     int max;     for (i=1;i<=n;i++)     {         for (k=1;k<=n;k++)            b[k]=0;         for (j=i;j<=n;j++)         {            for (k=1;k<=n;k++)            b[k]+=a[j][k];            max=f(n,b);         if (sum<max)         sum=max;        }     }     return sum;}int main(){    int i,j;    int n;    int a[150][150];    while (~scanf ("%d",&n))    {    for (i=1;i<=n;i++)        for (j=1;j<=n;j++)        scanf ("%d",&a[i][j]);        printf ("%d\n",sum (n,a));    }    return 0;}