POJ 1050:To the Max
To the Max
Time Limit: 1000MS | Memory Limit: 10000K | |
Total Submissions: 43241 | Accepted: 22934 |
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
题意是给定一个矩阵,求其子矩阵的最大和。
这题也是弄得相当郁闷,一开始暴力,结果预料之中的TLE。然后试了一下dp,结果还MLE。。。郁闷得不行。
然后看了别人的思路,发现可以二维变一维,想了想忽然恍然大悟。
将每一列的加起来,就是一维了。枚举不同行即可。之前怎么做的这次怎么求。
代码:
#include <iostream> #include <string> #include <cstring> #include <algorithm> #include <cmath> #pragma warning(disable:4996) using namespace std; int value[250][250]; int value2[250]; int dp[250]; int main() { //freopen("input.txt","r",stdin); //freopen("out.txt","w",stdout); int N,i,j,h,k,g,f; int ans=-100; scanf("%d",&N); memset(dp,0,sizeof(dp)); memset(value2,0,sizeof(value2)); for(i=1;i<=N;i++) { for(j=1;j<=N;j++) { scanf("%d",&value[i][j]); ans=max(ans,value[i][j]); } } for(i=1;i<=N;i++) { for(h=i;h<=N;h++) { for(k=1;k<=N;k++) { value2[k] += value[h][k]; dp[k] = max(dp[k-1]+value2[k],value2[k]); ans = max(ans,dp[k]); } memset(dp,0,sizeof(dp)); } memset(value2,0,sizeof(value2)); } cout<<ans<<endl; return 0; }
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