To the Max(矩阵压缩)

To the Max

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 20000/10000K (Java/Other)

Total Submission(s) : 2   Accepted Submission(s) : 2

Problem 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

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

8 0 -2

Sample Output

15

题解：dp问题善于把复杂问题简单化，过程异途同归；

代码：

 1 #include<stdio.h>
 2 #include<string.h>
 3 #define MAX(x,y)(x>y?x:y)
 4 const int INF=-0x3f3f3f3f;
 5 const int MAXN=110;
 6 int ans,N;
 7 void maxline(int *a){
 8     int sum=0;
 9     for(int i=1;i<=N;i++){
10     //    printf("%d ",a[i]);
11         if(sum>0)sum+=a[i];//保证sum+a[i]>a[i];
12         else sum=a[i];
13     //    printf("%d\n",sum);
14         ans=MAX(ans,sum);
15     }
16 }
17 int main(){
18     int s[MAXN],dp[MAXN],map[MAXN][MAXN];
19     while(~scanf("%d",&N)){
20         int temp;
21         ans=INF;
22         for(int i=1;i<=N;i++)
23             for(int j=1;j<=N;j++)
24                 scanf("%d",&map[i][j]);
25         for(int i=1;i<=N;i++){
26             for(int j=i;j<=N;j++){
27                 if(j-i)for(int k=1;k<=N;k++)
28                     map[i][k]+=map[j][k];
29                     maxline(map[i]);//这里是i代表每列从i行到j行的元素和
30             }
31         }
32         printf("%d\n",ans);
33     }
34     return 0;
35 }