ZOJ1074 (最大和子矩阵 DP)

F - 最大子矩阵和
Time Limit:1000MS Memory Limit:10000KB 64bit IO Format:%I64d & %I64uDescriptionGiven 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.InputThe 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].OutputOutput 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

题解：怎么说呢。。就是第一行的一个序列加到第二行，找到他们的和序列的最大子序列，然后他们的和序列再加第三行，再找，每加一次找一次，加到第n行。
然后从第二行开始，按照上面的进行，再从第三行开始..... 直到第n行....
每次都要更新他们的子矩阵的的最大值，用一个变量更新
不说了，上代码，最下面有输出截图

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
int n,sum,f,Max;
int b[20000];
int dp[150][150];
int main()
{
f=-100;
cin>>n;
for(int i=0; i<n; i++)
for(int j=0; j<n; j++)
{
cin>>dp[i][j];
}
for(int k=0; k<n; k++)
{
memset(b,0,sizeof(b));      // 一定每次记得清零
for(int i=k; i<n; i++)
{
for(int j=0; j<n; j++)
{
b[j]+=dp[i][j];
}
sum = Max =-100;         //赋很小的值
for(int i=0; i<n; i++)
{
if (sum<0)
sum = b[i];
else
sum += b[i];
if (Max < sum)
Max = sum;
}
if(f<Max)                  // 每次都要比较更新最大子矩阵和
f=Max;
}
}
cout<<f<<endl;
}