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POJ1050 To the Max 最大子矩阵

POJ1050 给定一个矩阵,求和最大的子矩阵. 将每一列的值进行累加,枚举起始行和结束行,然后就可以线性优化了 复杂度O(n^3) #include<cstdio> #include<cstdlib> #include<cstring> #include<cmath> #include<vector> #include<algorithm> using namespace std; const int N=301,M=301; co…

[POJ1050]To the Max

[POJ1050]To the Max 试题描述 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…

（线性dp 最大子段和 最大子矩阵和）POJ1050 To the Max

To the Max Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 54338   Accepted: 28752 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 wi…

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[POJ1050]To the Max (矩阵,最大连续子序列和)

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poj 1050 To the Max(最大子矩阵之和，基础DP题)

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[POJ1050] To the Max 及最大子段和与最大矩阵和的求解方法

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POJ 1050 To the Max 最大子矩阵和（二维的最大字段和）

传送门: http://poj.org/problem?id=1050 To the Max Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 52306   Accepted: 27646 Description Given a two-dimensional array of positive and negative integers, a sub-rectangle is any contiguous sub-arr…

POJ 1050 To the Max (最大子矩阵和)

题目链接 题意:给定N*N的矩阵,求该矩阵中和最大的子矩阵的和. 题解:把二维转化成一维,算下就好了. #include <cstdio> #include <cstring> #include <algorithm> using namespace std; typedef long long ll; ; ll ans=-,tmp,data[N][N],a[N]; ,ansj1=,ansi2=,ansj2=,tmp1=,tmp2=,n,m; int main() {…

HDU 1081 To the Max 最大子矩阵（动态规划求最大连续子序列和）

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. I…