POJ 1050 To the Max【DP】
阿新 • • 發佈:2019-01-07
To the Max
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.
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 the sum of the maximal sub-rectangle.
Time Limit: 1000MS | Memory Limit: 10000K |
Total Submissions: 48554 | Accepted: 25678 |
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
Output
Sample Input
4 0 -2 -7 0 9 2 -6 2 -4 1 -4 1 -1 8 0 -2
Sample Output
15
Source
做這題前,先弄清楚HDU1003這題。
然後再把二維壓縮為一維就可以了。
1,12,123,1234,2,23,234,3,34,4
按上面幾種情況壓縮,求出最大的即可。
AC程式碼:
/** * 行有餘力,則來刷題! * 部落格連結:http://blog.csdn.net/hurmishine * */ #include <iostream> #include <cstdio> #include <cstring> using namespace std; const int maxn=100+5; int a[maxn][maxn]; int dp[maxn]; int n; int getMax(int *a) { int maxx=a[0]; int sum=a[0]; for(int i=1;i<n;i++) { if(sum+a[i]>a[i]) sum+=a[i]; else sum=a[i]; if(sum>maxx) maxx=sum; } return maxx; } int main() { while(cin>>n) { for(int i=0;i<n;i++) { for(int j=0;j<n;j++) cin>>a[i][j]; } int p; int maxx=0; for(int i=0;i<n;i++) { memset(dp,0,sizeof(dp)); for(int j=i;j<n;j++) { p=0; for(int k=0;k<n;k++) { dp[p]+=a[j][k]; p++; } int ans=getMax(dp); if(ans>maxx) maxx=ans; } } cout<<maxx<<endl; } return 0; }