HDU 1081 To The Max (DP) 扩展最大子列和,求最大子矩阵和

To The Max

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3768    Accepted Submission(s): 1798


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
 
 
 
#include<stdio.h>
#include<string.h>
#define MAXN 110
int map[110][110];
/*//数组a,长度为n,的最大连续子段和*/
int MaxSubArray(int *a,int n)
{
int Max=-65535;
int i,tmp=0;
for(i=0;i<n;i++)
{
if(tmp>0)tmp+=a[i];
else tmp=a[i];
if(tmp>Max) Max=tmp;
}
return Max;
}
/*//求 n 行,m 列的矩阵的最大子矩阵和 */
int MaxSubMatrix(int n,int m)
{
int Max=-65535;
int i,j,k;
int sum;
int b[MAXN];
for(i=0;i<n;i++)
{
memset(b,0,sizeof(b));
for(j=i;j<n;j++)
{
for(k=0;k<m;k++) b[k]+=map[j][k];
sum=MaxSubArray(b,m);
if(sum>Max)Max=sum;
}
}
return Max;
}
int main()
{
int n;
int i,j;
while(scanf("%d",&n)!=EOF)
{
for(i=0;i<n;i++)
for(j=0;j<n;j++)
scanf("%d",&map[i][j]);
printf("%d\n",MaxSubMatrix(n,n));
}
return 0;
}

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