poj1050 To the Max(降维dp)
2024-10-21 09:36:49
To the Max
Time Limit: 1000MS | Memory Limit: 10000K | |
Total Submissions: 49351 | Accepted: 26142 |
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.
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
题意:
求矩阵中和最大的子矩阵。
思路:
把题目转化成一维的最大连续子段和。
枚举起始行i和终止行j,将i~j行的数对应相加,求解最大连续子段和。
代码:
#include<iostream>
using namespace std;
const int maxn = 105;
int grap[maxn][maxn], sum[maxn][maxn];
int main()
{
int n, tmp, ans=-0x3f3f3f3f;
cin>>n;
for(int i=1; i<=n; ++i)
{
for(int j=1; j<=n; ++j)
{
cin>>grap[i][j];
sum[i][j]=sum[i-1][j]+grap[i][j];
}
}
for(int i=1; i<=n; ++i)
{
for(int j=i; j<=n; ++j)
{
tmp=0;
for(int k=1; k<=n; ++k)
{
if(tmp<=0)
tmp=sum[j][k]-sum[i-1][k];
else
tmp+=sum[j][k]-sum[i-1][k];
ans=max(ans, tmp);
}
}
}
cout<<ans<<endl;
return 0;
}
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