poj 1050 To the Max 最大子矩阵和 经典dp
To the Max
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
Source
思路:经典dp(yan 教的)
#include<iostream> #include<cstdio> #include<cmath> #include<string> #include<queue> #include<algorithm> #include<stack> #include<cstring> #include<vector> #include<list> #include<set> #include<map> using namespace std; #define ll __int64 #define mod 1000000007 #define pi (4*atan(1.0)) const int N=1e3+10,M=1e6+10,inf=1e9+10; int a[N][N]; int sum[N][N]; int num[N]; int main() { int x,y,z,i,t; while(~scanf("%d",&x)) { memset(sum,0,sizeof(sum)); for(i=1;i<=x;i++) for(t=1;t<=x;t++) scanf("%d",&a[i][t]); for(i=1;i<=x;i++) for(t=1;t<=x;t++) sum[t][i]=sum[t-1][i]+a[t][i]; int maxx=-inf; for(i=1;i<=x;i++) { for(t=i;t<=x;t++) { for(int j=1;j<=x;j++) num[j]=sum[t][j]-sum[i-1][j]; int sum=0; for(int j=1;j<=x;j++) { sum+=num[j]; maxx=max(maxx,sum); if(sum<0) sum=0; } } } printf("%d\n",maxx); } return 0; }