To the Max POJ 1050（一維序列的變形）
阿新 • • 發佈：2018-12-26
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.
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
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
思路：題目看似是求最大子矩陣，思路和最大公共子序列一樣，a[i][j]代表了前i行的前j列的所有元素的和，然後直接套路，值得注意的是個人習慣，DP題目陣列下標從1開始
可以避免一些麻煩.
#include <cstdio> #include <cstring> #include <iostream> #include <algorithm> #include <vector> #include <queue> #include <set> #include <map> #include <string> #include <cmath> #include <cstdio> typedef long long ll; #define N 105 using namespace std; int a[N][N]; int main() { int i,j,k,n,t,sum,imax; while (scanf("%d",&n)!=EOF) { memset(a,0,sizeof(a)); for (i=1;i<=n;++i) { for (j=1;j<=n;++j) { scanf("%d",&t); a[i][j]=a[i-1][j]+t; } } imax=0; for (i=1;i<=n;++i) //第一行 { for (j=i;j<=n;++j) //第二行 { sum=0; for (k=1;k<=n;++k) { t=a[j][k]-a[i-1][k];//前j行前k列的和-前i-1行前k列的和 sum+=t; if (sum<0) sum=0; if (sum>imax) imax=sum; } } } printf("%d\n",imax); } return 0; }