hdu 1081 To The Max
时间：2014-04-27 19:54:26
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To The Max
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 7533 Accepted Submission(s): 3647
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.
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
动态规划求最大子矩阵和，把二维的转化为一维的就好做了。
1、首先看一位数组a[]求最大字段和，b记录以i结尾的数组最大字段和；
显然若b<0（此时存储的是以i-1为结尾的最大和）,有b=a[i]; 否则b+=a[i];
2、转化为一维的数组就是枚举某一连续几行，把每一列的值加起来存到一个元素中，
这些元素就构成了一个一维数组。
然后，按照一维数组求最大字段和的方法求解即可。
#include"stdio.h"
#include"math.h"
#include"string.h"
#define N 105
int Maxsum(int a[],int m)
{ //求一维数组的最大字段和
int i,s=0,max=a[0];
for(i=0;i<m;i++)
{
if(s>=0)
s+=a[i];
else
s=a[i];
if(s>max)
max=s;
}
return max;
}
int main()
{
int n,i,j,k;
int b[N];
int a[N][N];
while(scanf("%d",&n)!=-1)
{
for(i=0;i<n;i++)
for(j=0;j<n;j++)
scanf("%d",&a[i][j]);
int max=a[0][0];
for(i=0;i<n;i++)
{
memset(b,0,sizeof(b));
for(j=i;j<n;j++) //从某一行开始加到末行
{
for(k=0;k<n;k++) //求解相加过程中的最大字段和
b[k]+=a[j][k];
int t=Maxsum(b,n);
if(t>max)
max=t;
}
}
printf("%d\n",max);
}
return 0;
}
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