PKU 1050-To The Max(找矩形内元素最大和)
2015-08-20 09:07
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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.
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
题意:给一个N*N的矩形数组,求其中元素和最大的子矩形,输出最大值。
解析:因为数据比较小,可以枚举每个矩形,但是求矩形和时可以预处理一下。可以设置一个数组比如RecSum[i][j],i,j分别代表行,RecSum[i][j]表示右下角坐标为(i,j)左上角为(1,1)的矩形元素和,那么求某个矩形时,如左上角坐标为(upx,upy),右下角坐标为(lowx,lowy),则体积 V=RecSum[lowx][lowy]-RecSum[upx][lowy]-(RecSum[lowx][upy]-RecSum[upx][upy]);最后找最大值即可。
代码如下:
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
题意:给一个N*N的矩形数组,求其中元素和最大的子矩形,输出最大值。
解析:因为数据比较小,可以枚举每个矩形,但是求矩形和时可以预处理一下。可以设置一个数组比如RecSum[i][j],i,j分别代表行,RecSum[i][j]表示右下角坐标为(i,j)左上角为(1,1)的矩形元素和,那么求某个矩形时,如左上角坐标为(upx,upy),右下角坐标为(lowx,lowy),则体积 V=RecSum[lowx][lowy]-RecSum[upx][lowy]-(RecSum[lowx][upy]-RecSum[upx][upy]);最后找最大值即可。
代码如下:
#include<cstdio> #include<cstring> #include<string> #include<algorithm> #include<set> #include<map> #include<queue> #include<vector> #include<iterator> #include<utility> #include<sstream> #include<iostream> #include<cmath> #include<stack> using namespace std; const int INF=1000000007; const double eps=0.00000001; int N,elem[101][101]; int RowSum[101][101],RecSum[101][101]; inline void Get_RowSum() //处理每一行的前m个数的元素和 { memset(RowSum,0,sizeof(RowSum)); for(int x=1;x<=N;x++) for(int y=1;y<=N;y++) { if(y==1) RowSum[x][y]=elem[x][y]; else RowSum[x][y]=RowSum[x][y-1]+elem[x][y]; } } inline void Get_RecSum() //得到RecSum[][] { memset(RecSum,0,sizeof(RecSum)); for(int x=1;x<=N;x++) for(int y=1;y<=N;y++) { if(x==1) RecSum[x][y]=RowSum[x][y]; else RecSum[x][y]=RecSum[x-1][y]+RowSum[x][y]; } } inline int Get(int upx,int upy,int lowx,int lowy) { return RecSum[lowx][lowy]-RecSum[upx][lowy]-(RecSum[lowx][upy]-RecSum[upx][upy]); } int Cal(int row,int col) { int ret=-INF; for(int i=0;i+row<=N;i++) //枚举每个矩形 { for(int j=0;j+col<=N;j++) { int upx=i,upy=j,lowx=i+row,lowy=j+col; ret=max(ret,Get(upx,upy,lowx,lowy)); } } return ret; } int main() { while(cin>>N) { for(int i=1;i<=N;i++) for(int j=1;j<=N;j++) scanf("%d",&elem[i][j]); Get_RowSum(); Get_RecSum(); int ans=-INF; for(int row=1;row<=N;row++) // 枚举矩形大小 for(int col=1;col<=N;col++) ans=max(ans,Cal(row,col)); cout<<ans<<endl; } return 0; }
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