hdu 1081 (最大子矩陣和)dp To The Max
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.
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 題意:題目不難理解呀。就是給你一個nn的矩陣,讓你找出和最大的子矩陣來。 這種題原來做過,好久沒看了,忘了。。。。大體的思路就是先將原來矩陣每一行的數存到一個nn的字首和數組裡。然後控制列數,移動行數,這樣來找尋最大子矩陣。有的題解說這樣是二維壓縮成一維,我也不太明白。反正思路明白就好了。 程式碼如下:
#include<iostream> #include<cstdio> #include<cmath> #include<algorithm> #define inf 0x3f3f3f3f using namespace std; int a[101][101]; int n; int temp; int main() { while(scanf("%d",&n)!=EOF) { for(int i=1;i<=n;i++) { for(int j=1;j<=n;j++) { scanf("%d",&temp); a[i][j]=a[i][j-1]+temp; } } int ans=-inf; int res=0; for(int i=1;i<=n;i++) { for(int j=i;j<=n;j++) { for(int k=1,res=0;k<=n;k++)//每次行數從頭開始的時候就要將res更新為0 { res+=(a[k][j]-a[k][i-1]); if(res<0) res=0;//res小於0時,res就要更新為零!!! ans=max(ans,res); } } } printf("%d\n",ans); } return 0; }
努力加油a啊,(o)/~