To the Max

Description

Given a two-dimensional array of positive and negative integers, a sub-rectangle is any contiguous sub-array of size 1*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 * 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


最大字串和的升级
对于数列A[1->n]   dp[i]=dp[i-1]+a[i]  dp[i-1]>0  ||  dp[i]=a[i]  dp[i-1]<=0;
现在将矩阵建立与最大字串和相关联的模型。
矩阵压缩。滚动数组。

用s[k]表示第K列，第I行到第J行的和，把和看成一个数列中的一个AI即可。就转化成了最大字串和的问题。

#include"iostream"
#include"cstdio"
#include"cstring"
using namespace std;
const int ms=110;
int n,ans;
int matrix[ms][ms];
int s[ms];
int main()
{
    int i,j,k,t;
    while(scanf("%d",&n)!=EOF)
    {
        for(int i=1;i<=n;i++)
            for(j=1;j<=n;j++)
                scanf("%d",&matrix[i][j]);
        ans=-0x7fffffff;
        for(i=1;i<=n;i++)   //从第I行出发的子矩阵
        {
            memset(s,0,sizeof(s));
            for(j=i;j<=n;j++)   //到达第J行的子矩阵
            {
                t=0;
                for(k=1;k<=n;k++)//包含第K列 最大矩阵
                {
                    s[k]+=matrix[j][k];
                    if(t<=0)
                        t=s[k];
                    else
                        t+=s[k];
                    if(t>ans)
                        ans=t;
                        
                } 
            } 
        } 
        printf("%d\n",ans);
    }
    return 0;
}

接下来三维矩阵

同样的压缩思想。把立体——>平面——>直线

#include"iostream"
#include"cstdio"
#include"cstring"
#include"algorithm"
using namespace std;
const int ms=101;
int t,n,m;
int num[ms][ms][ms];
int submax(int a[ms]);
int submax2d(int a[][ms]);
int submax3d();
int main()
{
    int i,j,k;
    int test;
    cin>>test;
    while(test--)
    {
        cin>>t>>n>>m;
        for(k=1;k<=t;k++)
            for(i=1;i<=n;i++)
                for(j=1;j<=m;j++)
                    cin>>num[k][i][j];
    //    cout<<submax3d()<<endl;
        int ans=submax3d();
        cout<<ans<<endl;
    }
    return 0;
}
int submax(int a[ms])
{
    int i,pre=a[1],max=a[1];
    for(i=2;i<=m;i++)
    {
        if(pre>0)
            pre+=a[i];
        else
            pre=a[i];
        if(pre>max)
            max=pre;
    }
    return max;
}
int submax2d(int a[][ms])
{
    int b[ms];
    int i,j,k,max=a[1][1];
    for(i=1;i<=n;i++)
    {
        memset(b,0,sizeof(b));
        for(j=i;j<=n;j++)
        {
            for(k=1;k<=m;k++)
            {
                b[k]+=a[j][k];
            }
            int ff=submax(b);
            if(ff>max)
                max=ff;
        }
    }
    return max;
}
int submax3d()
{
    int a[ms][ms];
    int i,j,k,w;
    int max=num[1][1][1];
    for(i=1;i<=t;i++)
    {
        memset(a,0,sizeof(a));
        for(j=i;j<=t;j++)
        {
            for(k=1;k<=n;k++)
                for(w=1;w<=m;w++)
                    a[k][w]+=num[j][k][w];
        }
        int tt=submax2d(a);
        if(tt>max)
            max=tt;
    }
    return max;
}