poj1050 To the Max 最大子矩陣

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

40 -2 -7 0 9 2 -6 2-4 1 -4  1 -18  0 -2

Sample Output

15

最大子矩陣，首先一行數列很簡單求最大的子和，我們要把矩陣轉化成一行數列，就是從上向下在輸入的時候取和，map[i][j]表示在J列從上向下的數和，這樣就把一列轉化成了一個點，再用雙重，迴圈，任意i行j列開始的一排數的最大和，就是最終的最大和,其實，這就是，二維轉化成了一維，其實，我們擴充開來，如果是一個三維的呢，也就是一個長方體的數列，我們要找一個最大的小長方體，要求最大和，那麼，我們就可以這樣做，在數列中，我們可以把三維轉化成二維，二維轉化成一維，詳見以下代碼！

#include <iostream>#include <stdio.h>using namespace std;int map[105][105];int main(){    int n,i,j,k,sum,x,max;    while(scanf("%d",&n)!=EOF)    {        for(i=0;i<n;i++)            for(j=0;j<n;j++)            {                scanf("%d",&x);                map[i][j]=map[i-1][j]+x;            }            max=-0x4f4f4f4f;        for(i=0;i<n;i++)            for(j=i;j<n;j++)            {                sum=0;                for(k=0;k<n;k++)                {                    sum+=map[j][k]-map[i][k];                    if(sum<0)//小於0就相當於不用取了，直接去掉

                        sum=0;                    if(sum>max)                        max=sum;                }            }            printf("%d\n",max);    }    return 0;}

三維的最大子矩和

#include<string>#include<stdio.h>#include<stdlib.h>#include<string.h>#define M 101int t,n,m;int num[M][M][M];int submax(int a[M])//一維最大和{    int i,pre=a[1],max=0;    for(i=2;i<=m;i++)    {        if(a[i]+pre>a[i])            pre=a[i]+pre;        else        pre=a[i];        if(pre>max)        {            max=pre;        }    }    return max;}int submax2d(int a[][M])//二維最大和{    int b[M];    int i,j,k,max=0;    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];//二維壓成一維            }            int ff=submax(b);            if(ff>max)max=ff;        }    }    return max;}int submax3d()//三維最大和{    int a[M][M];    int i,j,k,w;    int max=0;    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];//三維壓成二維                }            int tt=submax2d(a);            if(tt>max)                max=tt;        }    }    return max;}int main (){    int i,j,k;    int test;    test=10;    scanf("%d",&test);    while(test--)    {       scanf("%d%d%d",&t,&n,&m);        for(k=1;k<=t;k++)            for(i=1;i<=n;i++)                for(j=1;j<=m;j++)                    scanf("%d",&num[k][i][j]);        printf("%d\n",submax3d());    }}