hdu 1081 To The Max ( 最大子矩阵 )
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To The Max
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 6612 Accepted Submission(s): 3159
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:
求矩阵里面的>=1*1规模的子矩阵的最大sum和。sum为子矩阵中各元素的和。
分析:
这题相当于是求二维的最大子串和,想想当初做hdu 1003 max sum一维的时候,就是看当前sum+第i个数,
如果sum<0,则sum=0并以第i+1个数为第一个数继续累加。每次为负就重复这个过程。
现在想想,其实就是一个状态转移的公式:
f[i]=max(0,f[i-1])+a[i]
至于这题,开始找最优子结构的是时候处理错了,开始想的是先找4个为第一层最小结构,然后再接着扩展。
但是这样很容易漏掉一些情况,且对数据处理也很麻烦,甚至到最后自己都搞晕了。
正确的思路是:
先把矩阵预处理。使得map[i][j]存的是第i行前j个数的和。
例如:样例中矩阵
0 -2 -7 0
9 2 -6 2
-4 1 -4 1
-1 8 0 -2
处理后变成:
0 -2 -9 -9
9 11 5 7
-4 -3 -7 -6
-1 7 7 5
三重for循环,sum+=map[k][j]-map[k][i-1]。 ( map[k][j]-map[k][i-1]表示第k行i列和j列之间的元素和)
注意k为最内层循环,每次都是确定两列后,将行边累加,边找maxsum。
列数的确定也要按一定的顺序,首先i固定在1,j从i到n。然后依次增大i...其实就是外面的两重循环。
这样就能找到遍历所有的子矩阵,找到最大的sum.
注意:max不能初始化为0,因为最大sum可能为负。
代码:
#include<cstdio>#include<iostream>#include<algorithm>#include<cstring>#define INF 0x3f3f3f3fusing namespace std;int map[101][101];int main(){ int tmp,i,j,k,n; while(scanf("%d",&n)!=EOF) { memset(map,0,sizeof(map)); for(i=1;i<=n;i++) { for(j=1;j<=n;j++) { scanf("%d",&tmp); map[i][j]=map[i][j-1]+tmp; } } int maxsum=-INF,sum; for(i=1;i<=n;i++) { for(j=i;j<=n;j++) { sum=0; for(k=1;k<=n;k++) { sum+=map[k][j]-map[k][i-1]; //printf("yes: %d\n",map[k][j]-map[k][i-1]); //printf("test: %d\n",sum); if(sum<0) sum=0; if(sum>maxsum) maxsum=sum; //printf("test: %d %d\n",sum,maxsum); } } } printf("%d\n",maxsum); } return 0;}
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