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

题解：dp问题善于把复杂问题简单化，过程异途同归；

代码：

 #include<stdio.h>

 #include<string.h>

 #define MAX(x,y)(x>y?x:y)

 const int INF=-0x3f3f3f3f;

 const int MAXN=;

 int ans,N;

 void maxline(int *a){

     int sum=;

     for(int i=;i<=N;i++){

     //    printf("%d ",a[i]);

         if(sum>)sum+=a[i];//保证sum+a[i]>a[i];

         else sum=a[i];

     //    printf("%d\n",sum);

         ans=MAX(ans,sum);

     }

 }

 int main(){

     int s[MAXN],dp[MAXN],map[MAXN][MAXN];

     while(~scanf("%d",&N)){

         int temp;

         ans=INF;

         for(int i=;i<=N;i++)

             for(int j=;j<=N;j++)

                 scanf("%d",&map[i][j]);

         for(int i=;i<=N;i++){

             for(int j=i;j<=N;j++){

                 if(j-i)for(int k=;k<=N;k++)

                     map[i][k]+=map[j][k];

                     maxline(map[i]);//这里是i代表每列从i行到j行的元素和

             }

         }

         printf("%d\n",ans);

     }

     return ;

 }

巴特西

To the Max(矩阵压缩)

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

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

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