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
最大字串和的升级
对于数列A[1->n] dp[i]=dp[i-1]+a[i] dp[i-1]>0 || dp[i]=a[i] dp[i-1]<=0;
现在将矩阵建立与最大字串和相关联的模型。
矩阵压缩。滚动数组。
用s[k]表示第K列，第I行到第J行的和，把和看成一个数列中的一个AI即可。就转化成了最大字串和的问题。
1 #include"iostream" 2 #include"cstdio" 3 #include"cstring" 4 using namespace std; 5 const int ms=110; 6 int n,ans; 7 int matrix[ms][ms]; 8 int s[ms]; 9 int main() 10 { 11 int i,j,k,t; 12 while(scanf("%d",&n)!=EOF) 13 { 14 for(int i=1;i<=n;i++) 15 for(j=1;j<=n;j++) 16 scanf("%d",&matrix[i][j]); 17 ans=-0x7fffffff; 18 for(i=1;i<=n;i++) //从第I行出发的子矩阵 19 { 20 memset(s,0,sizeof(s)); 21 for(j=i;j<=n;j++) //到达第J行的子矩阵 22 { 23 t=0; 24 for(k=1;k<=n;k++)//包含第K列 最大矩阵 25 { 26 s[k]+=matrix[j][k]; 27 if(t<=0) 28 t=s[k]; 29 else 30 t+=s[k]; 31 if(t>ans) 32 ans=t; 33 34 } 35 } 36 } 37 printf("%d ",ans); 38 } 39 return 0; 40 }
接下来三维矩阵
同样的压缩思想。把立体——>平面——>直线
1 #include"iostream" 2 #include"cstdio" 3 #include"cstring" 4 #include"algorithm" 5 using namespace std; 6 const int ms=101; 7 int t,n,m; 8 int num[ms][ms][ms]; 9 int submax(int a[ms]); 10 int submax2d(int a[][ms]); 11 int submax3d(); 12 int main() 13 { 14 int i,j,k; 15 int test; 16 cin>>test; 17 while(test--) 18 { 19 cin>>t>>n>>m; 20 for(k=1;k<=t;k++) 21 for(i=1;i<=n;i++) 22 for(j=1;j<=m;j++) 23 cin>>num[k][i][j]; 24 // cout<<submax3d()<<endl; 25 int ans=submax3d(); 26 cout<<ans<<endl; 27 } 28 return 0; 29 } 30 int submax(int a[ms]) 31 { 32 int i,pre=a[1],max=a[1]; 33 for(i=2;i<=m;i++) 34 { 35 if(pre>0) 36 pre+=a[i]; 37 else 38 pre=a[i]; 39 if(pre>max) 40 max=pre; 41 } 42 return max; 43 } 44 int submax2d(int a[][ms]) 45 { 46 int b[ms]; 47 int i,j,k,max=a[1][1]; 48 for(i=1;i<=n;i++) 49 { 50 memset(b,0,sizeof(b)); 51 for(j=i;j<=n;j++) 52 { 53 for(k=1;k<=m;k++) 54 { 55 b[k]+=a[j][k]; 56 } 57 int ff=submax(b); 58 if(ff>max) 59 max=ff; 60 } 61 } 62 return max; 63 } 64 int submax3d() 65 { 66 int a[ms][ms]; 67 int i,j,k,w; 68 int max=num[1][1][1]; 69 for(i=1;i<=t;i++) 70 { 71 memset(a,0,sizeof(a)); 72 for(j=i;j<=t;j++) 73 { 74 for(k=1;k<=n;k++) 75 for(w=1;w<=m;w++) 76 a[k][w]+=num[j][k][w]; 77 } 78 int tt=submax2d(a); 79 if(tt>max) 80 max=tt; 81 } 82 return max; 83 }