poj1050

经典的最大子矩阵和。。

转化成最大子段和

（通过枚举列的长度，然后对于选定的那一列做一次的最长子段和）最后选取最大就是答案。。

/*
 State:Accepted
 Time:2013.3.1
*/

#include <iostream>
#include <cstdlib>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <fstream>
#include <string>
#include <cmath>
using namespace std;
int n , a[150][150] ,f[200], sumrow[200][200] , ans ;

void init(){      
      scanf("%d",&n);
      for (int i = 1; i <= n; ++i)
          for (int j = 1; j <= n; ++j){
              scanf("%d",&a[i][j]);
              sumrow[i][j] = sumrow[i - 1][j] + a[i][j];
          }
      
}

void dp(){
     int nowv;
     ans = -1000000;
     for (int l = 1; l <= n; ++l)
        for (int r = l; r <= n; ++r){
            for (int i = 1; i <= n; ++i)  f[i] = -1000000;
            for (int i = 1; i <= n; ++i){
               nowv = sumrow[r][i] - sumrow[l - 1][i];
               f[i] = max(nowv , f[i -1] + nowv);
               ans  = max(ans , f[i]);
            }
        }
     printf("%d\n",ans);
     
}
int main(){
      freopen("poj1050.in","r",stdin);
      freopen("poj1050.out","w",stdout);
      init();
      dp();
      fclose(stdin);
      fclose(stdout);
}