sdut1359 求最大和子矩阵

题目描述

Given a two-dimensional array of positive and negative integers, a sub-rectangle is any contiguous sub-array of size 1 x 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.

输入

The input consists of an N x 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].

输出

Output the sum of the maximal sub-rectangle.

示例输入

4
0 -2 -7 0 9 2 -6 2
-4 1 -4 1 -1
8 0 -2

示例输出

 1 #include<stdio.h>
 2 #include<string.h>
 3 int dp(int a[],int n)
 4 {
 5     int i,f[200];
 6     int max=-200000000;
 7     f[1]=a[1];
 8     for(i=2; i<=n; i++)
 9     {
10         if(f[i-1]>0)
11             f[i]=f[i-1]+a[i];
12         else f[i]=a[i];
13         if(max<f[i])
14             max=f[i];
15     }
16     return max;
17 }
18 
19 int main()
20 {
21     int i,n,sum1,j;
22     int a[200][200],k;
23     int sum[200],max;
24     while(~scanf("%d",&n))
25     {
26         for(i=1; i<=n; i++)
27             for(j=1; j<=n; j++)
28                 scanf("%d",&a[i][j]);
29         max=-200000000;
30         for(i=1; i<=n; i++)
31         {
32             memset(sum,0,sizeof(sum));
33             for(j=i; j<=n; j++)
34             {
35                 for(k=1; k<=n; k++)
36                     sum[k]+=a[j][k];
37                 sum1=dp(sum,n);
38                 if(sum1<0)sum1=0;
39                 if(sum1>max)max=sum1;
40             }
41         }
42         printf("%d
",max);
43     }
44 
45     return 0;
46 }

 1 #include<stdio.h>
 2 #include<string.h>
 3 #define MAXN 105
 4 int main()
 5 {
 6     int i,j,k,n,t,sum,max;
 7     int a[MAXN][MAXN];
 8     while (scanf("%d",&n)!=EOF)
 9     {
10         memset(a,0,sizeof(a));
11         for (i=1; i<=n; ++i)
12         {
13             for (j=1; j<=n; ++j)
14             {
15                 scanf("%d",&t);
16                 a[i][j]=a[i-1][j]+t;
17             }
18         }
19         max=0;
20         for (i=1; i<=n; ++i)
21         {
22             for (j=i; j<=n; ++j)
23             {
24                 sum=0;
25                 for (k=1; k<=n; ++k)
26                 {
27                     t=a[j][k]-a[i-1][k];
28                     sum+=t;
29                     if (sum<0) sum=0;
30                     if (sum>max) max=sum;
31                 }
32             }
33         }
34         printf("%d
",max);
35     }
36     return 0;
37 }