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HDU 1081 To The Max
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
-18 0 -2
Sample Output
15
这道题是求大矩阵中小矩形的和的最大值
可以先算出num[i][j]代表第i行前面j列的值
然后相当于固定一列i,j从1~i中变化,K代表行从1~n开始循环,相当于能求出固定两行之间
的矩形的最大值(当然也可以看成求子序列的最大值了,因为每行可以的和可以看成一个数,就相当于压缩成了一维)
每次找到最大值更新结果就可以了
一维的最大连续子序列的递推公式:
f[i]=max{f[i-1]+a[i],a[i]} (以i结尾的最大连续子序列)
1 #include<stdio.h> 2 #include<string.h> 3 #include<stdlib.h> 4 #include<algorithm> 5 using namespace std; 6 int num[101][101]; 7 int ans; 8 int main() 9 {10 int n,i,j,k,a;11 while(scanf("%d",&n)!=EOF)12 {13 memset(num,0,sizeof(num));14 for(i=1;i<=n;i++){15 for(j=1;j<=n;j++){16 scanf("%d",&a);17 num[i][j]=num[i][j-1]+a;18 }19 }20 int maxn=-0x3fffffff;21 for(i=1;i<=n;i++)22 {23 for(j=1;j<=i;j++)24 {25 ans=-1;26 for(k=1;k<=n;k++)27 {28 if(ans>0)29 ans+=num[k][i]-num[k][j-1];30 else31 ans=num[k][i]-num[k][j-1];32 if(ans>maxn)33 maxn=ans;34 }35 }36 }37 printf("%d\n",maxn);38 }39 return 0;40 }
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