To the Max Time Limit: 1000MSMemory Limit: 10000K Total Submissions: 38573Accepted: 20350 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 t…

To The Max Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 11685    Accepted Submission(s): 5649 Problem Description Given a two-dimensional array of positive and negative integers, a sub-recta…

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…

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() {…

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…

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…

直接暴力枚举所有子矩形至少需要O(n^4)的复杂度,显然这不是一个合理的解决方法. 上述方案忽略了矩形之间的联系,进行了过多不必要的计算. 实际上如果固定矩形的左右边界,则底边在i行的矩形内数值之和与底边在i-1行的矩形的关系为 f[i] = s[i] + max(0, f[i - 1]), s[i]表示当前行对应的数值之和. 本题枚举是切入口,通过枚举把所有矩形分成左右边界固定的矩形族,而后再进行动态规划,可降低复杂度至O(n^3). acm.hdu.edu.cn/showproblem.ph…

To the Max   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…

http://poj.org/problem?id=1050 我们已经知道求最大子段和的dp算法 参考here  也可参考编程之美有关最大子矩阵和部分. 然后将这个扩大到二维就是这道题.顺便说一下,有时候不要把问题想复杂了,有些问题只能靠暴力求解,而这道题是暴力加算法. 在这个题中,我们可以把二维压缩到一维然后求解最大子段.我们先枚举所求矩阵的起点行和结束行,然后把每一列的数据之和求出,用这些数据和就构造出一个一维的数组(代码中我没有明确表示出这个数组),然后用最大子段和的dp算法求解. 关于二…

小故事: A * "1+1+1+1+1+1+1+1 =?" * A : "上面等式的值是多少" B : *计算* "8!" A *在上面等式的左边写上 "1+" * A : "此时等式的值为多少" B : *quickly* "9!" A : "你怎么这么快就知道答案了" A : "只要在8的基础上加1就行了" A : "所以你不用重新计…