poj1050--最大子序列和

http://acm.pku.edu.cn/JudgeOnline/problem?id=1050

                                                                                             To the Max

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 21973		Accepted: 11400

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

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

Sample Output

15

与hdu1003不同的题，但却是与1003相同的思想，具体代码如下：

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
 int n,a[101][101];
 int s[101];
int ma(int *p){  ————最大子序列的和，并返回最大值
    int i,max=-(1<<30),b;
    b=p[1];
    for(i=2;i<=n;i++)
    {
    if(b<0)b=p[i];
    else b=b+p[i];
    if(max<b)max=b;
}
    return max;   
}

int main(){
    int i,j,l,k,max=-(1<<30),t;
    scanf("%d",&n);
    for(i=1;i<=n;i++)
    for(j=1;j<=n;j++)
    scanf("%d",&a[i][j]);
    memset(s,0,sizeof(s));
    for(l=1;l<=n;l++) —————代表结束行
    for(i=1;i<=n;i++)——————代表起始行
    { for(j=i;j<=l;j++)
      for(k=1;k<=n;k++)
      s[k]+=a[j][k];——————s[k]中存第k列从i到l的所有数的和
      t=ma(s);
      if(max<t)max=t;                     
        memset(s,0,sizeof(s));
                     }
                     printf("%d\n",max);
                     return 0;
                     }

经验总结：编程过程中，注意运用化未知为已知的数学思维。 

本题虽然为矩阵，但思想就是将它转化为一列数，然后再求最大子序列和。但如何转化？

7 -8 9

-4 5 6

1 2 -3

主要是将同一列中的若干数合并。比如，从第一行开始，到第2行结束，每一列的和组成的序列为：

3 -3 15

然后求此序列的最大子序列和。求出后与max比较，最后输出的一定是最大矩阵和。

除了按照程序中按初始位置和结束位置枚举外，还可以枚举每一列中的元素个数和起始位置写循环。