poj 1050 To the Max(最大子矩阵权值) - c++编程基础

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 38220		Accepted: 20161

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

Source

Greater New York 2001

题意：

给你一个数字矩阵。要你找出一个子矩阵。使得该矩阵的和比任意其它子矩阵的和都大。求出最大值。

思路：

对于一维的最大连续和肯定很简单。dp[i]=max(arr[i],dp[i-1]+arr[i])。dp[i]表示最大连续和包含arr[i]的最大值。

那么二维就推广下。枚举子矩阵的上下边。把两边间的数字压缩成一条线就可以按一维处理了。

详细见代码：

#include
  
   
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             //#pragma comment(linker,"/STACK:1024000000,1024000000") using namespace std; const int INF=0x3f3f3f3f; const double eps=1e-8; const double PI=acos(-1.0); const int maxn=100010; //typedef __int64 ll; int sum[150][150],dp[150]; int main() { int data,n,i,j,k,ans,tp; memset(sum,0,sizeof sum); dp[0]=0; while(~scanf("%d",&n)) { ans=-INF; for(i=1;i<=n;i++) { for(j=1;j<=n;j++) { scanf("%d",&data); sum[j][i]=sum[j][i-1]+data; } } for(i=1;i<=n;i++) { for(j=1;j<=i;j++) { for(k=1;k<=n;k++) { tp=sum[k][i]-sum[k][i-j-1]; dp[k]=max(tp,dp[k-1]+tp); ans=max(dp[k],ans); } } } printf("%d\n",ans); } return 0; }