sdut1359 求最大和子矩阵

题目描述

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.

输入

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].

输出

Output the sum of the maximal sub-rectangle.

示例输入

4
0 -2 -7 0 9 2 -6 2
-4 1 -4 1 -1
8 0 -2

示例输出

15

#include<stdio.h>
#include<string.h>
int dp(int a[],int n)
{
    int i,f[200];
    int max=-200000000;
    f[1]=a[1];
    for(i=2; i<=n; i++)
    {
        if(f[i-1]>0)
            f[i]=f[i-1]+a[i];
        else f[i]=a[i];
        if(max<f[i])
            max=f[i];
    }
    return max;
}

int main()
{
    int i,n,sum1,j;
    int a[200][200],k;
    int sum[200],max;
    while(~scanf("%d",&n))
    {
        for(i=1; i<=n; i++)
            for(j=1; j<=n; j++)
                scanf("%d",&a[i][j]);
        max=-200000000;
        for(i=1; i<=n; i++)
        {
            memset(sum,0,sizeof(sum));
            for(j=i; j<=n; j++)
            {
                for(k=1; k<=n; k++)
                    sum[k]+=a[j][k];
                sum1=dp(sum,n);
                if(sum1<0)sum1=0;
                if(sum1>max)max=sum1;
            }
        }
        printf("%d
",max);
    }

    return 0;
}

#include<stdio.h>
#include<string.h>
#define MAXN 105
int main()
{
    int i,j,k,n,t,sum,max;
    int a[MAXN][MAXN];
    while (scanf("%d",&n)!=EOF)
    {
        memset(a,0,sizeof(a));
        for (i=1; i<=n; ++i)
        {
            for (j=1; j<=n; ++j)
            {
                scanf("%d",&t);
                a[i][j]=a[i-1][j]+t;
            }
        }
        max=0;
        for (i=1; i<=n; ++i)
        {
            for (j=i; j<=n; ++j)
            {
                sum=0;
                for (k=1; k<=n; ++k)
                {
                    t=a[j][k]-a[i-1][k];
                    sum+=t;
                    if (sum<0) sum=0;
                    if (sum>max) max=sum;
                }
            }
        }
        printf("%d
",max);
    }
    return 0;
}