题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1081
To The Max
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 8839 Accepted Submission(s):
4281
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
4
0 -2 -7 0 9 2 -6 2
-4 1 -4 1 -1
8 0 -2
Sample Output
15
题目大意:在一个数的矩阵里面找一个和最大矩阵。
题目思路:化二维为一维,转化成1003即可。先求一列一列的和,这样就转化成了一维了。举个例子,我先计算第一行得到dp[0]=0dp[1]=-2,dp[2]=-7,dp[3]=0;这些就是所谓的和,然后在用1003的方法做。那么第二行,就会更新,dp[0]=9,dp[1]=0,dp[2]=-13,dp[3]=2;这些依旧是和,这也就是转换成了一维数组,用1003的方法解决,以此类推~~~
详见代码。
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 5 using namespace std; 6 7 int dp[110]; 8 int num[110][110]; 9 10 int main () 11 { 12 int n; 13 while (~scanf("%d",&n)) 14 { 15 for (int i=0; i<n; i++) 16 for (int j=0; j<n; j++) 17 scanf("%d",&num[i][j]); 18 int ans=-99999; 19 for (int i=0; i<n; i++) 20 { 21 memset(dp,0,sizeof(dp)); 22 for (int j=i; j<n; j++) 23 { 24 int Max=-1; 25 for (int k=0; k<n; k++) 26 { 27 dp[k]+=num[j][k]; //先计算出所有的dp和 28 } 29 for (int k=0; k<n; k++) //1003的做法,代码类似 30 { 31 if (Max+dp[k]<dp[k]) 32 Max=dp[k]; 33 else 34 Max=Max+dp[k]; 35 if (ans<Max) //不断更新最大值 36 { 37 ans=Max; 38 //cout<<j<<" "<<k<<endl; 39 } 40 } 41 } 42 } 43 printf ("%d ",ans); 44 } 45 return 0; 46 }