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 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…

题目链接 Problem - 1081 题意 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…

题目:http://acm.hdu.edu.cn/showproblem.php?pid=1081 自己真够垃圾的,明明做过一维的这种题,但遇到二维的这种题目,竟然不会了,我也是服了(ps:猪啊). 最终还是看了题解. 代码如下: #include <iostream> #include <stdio.h> #include <string.h> #include <algorithm> #define inf 0x3f3f3f3f using namesp…

F - 最大子矩形 Time Limit:1000MS Memory Limit:10000KB 64bit IO Format:%I64d & %I64u Submit Status 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 withi…

最大字段和题型,推荐做题顺序: HDU1003    HDU1024     HDU1081  zoj2975  zoj2067 #include<cstdio> #include<cstdlib> #include<iostream> #include<memory.h> using namespace std; int n,a[110][110]; int y[110],maxp[110]; int Max; int main() { int i,j,k…

最大字段和题型,推荐做题顺序: HDU1003 HDU1024 HDU1081  ZOJ2975 ZOJ2067 #include<cstdio> #include<cstdlib> #include<iostream> #include<memory.h> using namespace std; ][]; ],maxp[]; int Max; int main() { int i,j,k; while(~scanf("%d",&…

今天查看Kafka 0.10.0的官方文档,发现了这样一句话:Configuration parameter replica.lag.max.messages was removed. Partition leaders will no longer consider the number of lagging messages when deciding which replicas are in sync. 即replica.lag.max.messages参数被正式地移除了,现在topic…

转载http://blog.csdn.net/Shayabean_/article/details/44885917博客 先说说基数排序的思想: 基数排序是非比较型的排序算法,其原理是将整数按位数切割成不同的数字,然后按每个位数分别比较. 将所有待比较数值(正整数)统一为同样的数位长度,数位较短的数前面补零.然后,从最低位开始,依次进行一次排序.在每一次排序中,按照当前位把数组元素放到对应 的桶当中,然后把桶0到桶9中的元素按先进先出的方式放回数组中.这样从最低位排序一直到最高位排序完成以后,…

Given a non-empty 2D matrix matrix and an integer k, find the max sum of a rectangle in the matrix such that its sum is no larger than k. Example: Given matrix = [ [1, 0, 1], [0, -2, 3] ] k = 2 The answer is 2. Because the sum of rectangle [[0, 1], […