POJ 1050 To the Max

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

题意 
求最大子矩阵和。

分析
一维最大连续子段和的扩展。首先输入遍历时可以找出max（每行的最大连续子段和，相当于宽度为一的子矩阵），
然后遍历，把第i行后的各行对应列的元素加到第i行的对应列元素，每加一行，就求一次最大字段和，
这样就把子矩阵的多行压缩为一行了，变成一行了之后就是最大字段和了！巧妙！

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<algorithm>
#include<cstring>
#include <queue>
#include <vector>
#include<bitset>
#include<map>
#include<deque>
using namespace std;
typedef long long LL;
const int maxn = 1e4+5;
const int mod = 77200211+233;
typedef pair<int,int> pii;
#define X first
#define Y second
#define pb push_back
//#define mp make_pair
#define ms(a,b) memset(a,b,sizeof(a))
const int inf = 0x3f3f3f3f;
#define lson l,m,2*rt
#define rson m+1,r,2*rt+1
typedef long long ll;
#define N 100010
int a[105][105];
int main(){
#ifdef LOCAL
    freopen("in.txt","r",stdin);
#endif // LOCAL
    int n,tmp;
    scanf("%d",&n);
    int ans = -inf;
    for(int i=1;i<=n;i++){
        tmp=0;
        for(int j=1;j<=n;j++){
            scanf("%d",&a[i][j]);
            if(tmp>0) tmp+=a[i][j];
            else tmp=a[i][j];
            ans = max(ans,tmp);
        }

    }

    for(int i=1;i<n;i++){
        for(int j=i+1;j<=n;j++){
            tmp=0;
            for(int k=1;k<=n;k++){
                a[i][k]+=a[j][k];
                if(tmp>0) tmp+= a[i][k];
                else tmp = a[i][k];
                ans = max(ans,tmp);
            }
        }
    }
    cout<<ans<<endl;
    return 0;
}

顺便给出求最大子段和的代码和思想：

int maxsum(int x[],int n)  
{  
   int i,b = 0,k = -10000000;  
   
   for(i = 0 ; i < n ; ++i)  
   {  
       if(b > 0) b += x[i];//如果累加和是正数，则继续加   
       /*  
          如果b <= 0,那么一定有x[i-1]<0,x[i]待定，那么如果x[i]>= 0时， 
          b=x[i]是理所当然的；如果x[i]<0呢？b=x[i]合适吗？答案是合适。 
          因为下一次循环b依然小于0，肯定可以找到一个大于0的数     
            
          还有一个问题：b = x[i],那不就想当然把刚才那个字段全部舍弃了吗？ 
          如果刚才那个子段的子段（前几个为负数）大于0呢？但这是不可能的。 
          因为一个字段的第一个数一定是个正数，因为如果第一个数是负数， 
          那么b<0，会执行else，直到有个正数出现，才会开始一个子段的累加  
       */  
           
       else  b = x[i];//如果累加和是负数了，就把这个值赋值给b    
               
       if(b > k) k = b;//更新最大字段和   
   }  
          
   return k;   
}