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hdu-------1081To The Max
To The Max
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 7681 Accepted Submission(s): 3724
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
Source
Greater New York 2001
这道题个人觉得是到灰常好的题目,适合不同层次的人做,求的是最大连续子矩阵和....
方法一:
简单的模拟即可。首先求出每个阶段只和,然后得到这个状态,然后进行矩阵规划,求最大值即可
代码浅显易懂:
1 #include<cstdio> 2 #include<cstring> 3 #include<cstdlib> 4 #include<iostream> 5 using namespace std; 6 int arr[101][101]; 7 int sum[101][101]; 8 int main() 9 { 10 int nn,i,j,ans,temp; 11 // freopen("test.in","r",stdin); 12 while(scanf("%d",&nn)!=EOF) 13 { 14 ans=0; 15 memset(sum,0,sizeof(sum)); 16 for(i=1;i<=nn;i++) 17 { 18 for(j=1;j<=nn;j++) 19 { 20 scanf("%d",&arr[i][j]); 21 sum[i][j]+=arr[i][j]+sum[i][j-1]; //求出每一层逐步之和 22 } 23 } 24 for(i=2;i<=nn;i++) 25 { 26 for(j=1;j<=nn;j++) 27 { 28 sum[i][j]+=sum[i-1][j]; //在和的基础上,逐步求出最大和 29 } 30 } 31 ans=0; 32 for(int rr=1;rr<=nn;rr++) 33 { 34 for(int cc=1;cc<=nn;cc++) 35 { 36 for(i=rr;i<=nn;i++) 37 { 38 for(j=cc;j<=nn;j++) 39 { 40 temp=sum[i][j]-sum[i][cc-1]-sum[rr-1][j]+sum[rr-1][cc-1]; 41 if(ans<temp) ans=temp; 42 } 43 } 44 } 45 } 46 printf("%d\n",ans); 47 } 48 return 0; 49 }
方法二:
采用状态压缩的方式进行DP求解最大值......!
代码:
1 /*基于一串连续数字进行求解的扩展*/ 2 /*coder Gxjun 1081*/ 3 #include<iostream> 4 #include<cstdio> 5 #include<cstring> 6 #include<cstdlib> 7 using namespace std; 8 const int nn=101; 9 int arr[nn][nn],sum[nn][nn]; 10 int main() 11 { 12 int n,i,j,maxc,res,ans; 13 // freopen("test.in","r",stdin); 14 while(scanf("%d",&n)!=EOF) 15 { 16 memset(sum,0,sizeof(sum)); 17 for(i=1;i<=n;i++) 18 for(j=1;j<=n;j++) 19 { 20 scanf("%d",&arr[i][j]); 21 sum[j][i]=sum[j][i-1]+arr[i][j]; 22 } 23 ans=0; 24 for(i=1;i<=n;i++){ 25 26 for(j=i;j<=n;j++) 27 { 28 res=maxc=0; 29 for(int k=1;k<=n;k++) 30 { 31 maxc+=sum[k][j]-sum[k][i-1]; 32 if(maxc<=0) maxc=0; 33 else 34 if(maxc>res) res=maxc; 35 } 36 if(ans<res) ans=res; 37 } 38 } 39 printf("%d\n",ans); 40 } 41 return 0; 42 }
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