先上题目:
To The Max
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 7177 Accepted Submission(s): 3470
Problem Description
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.
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 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].
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
二维版的求区间最值。
如果是一维版的话状态转移方程:
dp[i]=w[i] ans<=0
=ans+w[i] ans>0 其中ans代表当前位置i往前连续的一段的区域最值
而maxn=max(ans,maxn) maxn代表总体最值
对于二维的时候,其实和一维大致相同。这里需要在一个方向上(假设是行数增加的方向)枚举区间,然后在另一个方向上(列数增加的方向)上进行想一维的时候那样的操作。这样做的涵义其实是在某一个确定的区间上,求这个区间上的最值。
详细需要看代码。
分析如图:
上代码:
1 #include <cstdio> 2 #include <cstring> 3 #define MAX 101 4 #define INF ~(1<<31) 5 using namespace std; 6 7 int s[MAX][MAX]; 8 int h[MAX][MAX]; 9 int dp[MAX][MAX]; 10 11 int main() 12 { 13 int n; 14 int ans,maxn; 15 while(scanf("%d",&n)!=EOF) 16 { 17 memset(s,0,sizeof(s)); 18 memset(h,0,sizeof(h)); 19 for(int i=1; i<=n; i++) 20 { 21 for(int j=1; j<=n; j++) 22 { 23 scanf("%d",&s[i][j]); 24 h[i][j]=h[i][j-1]+s[i][j]; 25 } 26 } 27 maxn=-INF; 28 for(int i=1; i<=n; i++) 29 { 30 for(int j=1; j<=i; j++) 31 { 32 ans=-1; 33 for(int k=1; k<=n; k++) 34 { 35 if(ans>0) ans+=h[k][i]-h[k][j-1]; 36 else ans=h[k][i]-h[k][j-1]; 37 maxn=ans>maxn ? ans : maxn; 38 } 39 } 40 } 41 printf("%d ",maxn); 42 } 43 return 0; 44 }