hdu 1081 To The Max(二维最大连续和)
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To The Max
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 8735 Accepted Submission(s): 4234
Problem Description
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.
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
40 -2 -7 0 9 2 -6 2-4 1 -4 1 -18 0 -2
Sample Output
15
Source
Greater New York 2001
题意:
求出最大子矩阵的和
题解:
sum[i][j]表示第i行前j个元素的和,这样第i列到第j列之间的元素和就可以通过sum[k][j]-sum[k][i-1]算出来了,并看成一个元素,这样就成了n个元素的最大连续和(杭电1003题),枚举i,j就可以算出最大值。
代码:
#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>using namespace std;int sum[110][110];int main(){ int n; while(~scanf("%d",&n)) { int a,ans=-99999999; // memset(sum,0,sizeof(sum)); for(int i=1;i<=n;i++) { for(int j=1;j<=n;j++) { scanf("%d",&a); sum[i][j]=sum[i][j-1]+a;//第i行前j个元素之和 } } for(int i=1;i<=n;i++) { for(int j=i;j<=n;j++) { int temp=0; for(int k=1;k<=n;k++)//第i列到第j列的情况,最大连续和算法 { temp+=(sum[k][j]-sum[k][i-1]); if(temp>ans) ans=temp; if(temp<0) temp=0; } } } printf("%d\n",ans); } return 0;}
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