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], […

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], […

/* * 363. Max Sum of Rectangle No Larger Than K * 2016-7-15 by Mingyang */ public int maxSumSubmatrix(int[][] matrix, int target) { int row = matrix.length; if(row==0)return 0; int col = matrix[0].length; int m = Math.min(row,col); int n = Math.max(r…

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: Input: matrix = [[1,0,1],[0,-2,3]], k = 2 Output: 2 Explanation: Because the sum of rectangle [[0, 1],…

作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 题目地址: https://leetcode.com/problems/max-sum-of-rectangle-no-larger-than-k/description/ 题目描述: Given a non-empty 2D matrix matrix and an integer k, find the max sum of a rectangle in the matrix su…

题目描述: 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. 解题思路: 根据题意,寻找二维数组中所有可以组成的矩形中面积不超过k的最大值,所以必须要求出可能组成的矩形的面积并与k比较求出最终结果.这里为了最终不超时,可以在一下方面进行优化: 1.设置一个数组比较当前列(或…

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: Input: matrix = [[1,0,1],[0,-2,3]], k = 2 Output: 2 Explanation: Because the sum of rectangle [[0, 1],…

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 = 2The answer is 2. Because the sum of rectangle [[0, 1], [-2…

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], […

题目链接 链接:https://leetcode.com/problems/max-sum-of-rectangle-no-larger-than-k/description/ 题解&代码 1.暴力枚举所有的情况,时间复杂度O(n^2*m^2),实际耗时759 ms class Solution { public: int maxSumSubmatrix(vector<vector<int> >& matrix, int k) { int n=matrix.size…

Leetcode363 思路: 一种naive的算法就是枚举每个矩形块, 时间复杂度为O((mn)^2), 可以做少许优化时间复杂度可以降低到O(mnnlogm), 其中m为行数, n为列数. 先求出任意两列之间的所有数的和, 然后再枚举任意两行之间的和, 而我们优化的地方就在后者. 我们用s[x]来表示第x行从a列到b列的和. 遍历一遍从第0行到最后一行的求和数组, 并依次将其放到二叉搜索树中, 这样当我们知道了从第0行到当前行的和的值之后, 我们就可以用lower_bound在O(log n…

leetcode面试准备: Maximal Rectangle 1 题目 Given a 2D binary matrix filled with 0's and 1's, find the largest rectangle containing all ones and return its area. 接口: int maximalRectangle(char[][] matrix) 2 思路 这是一道非常综合的题目,要求在0-1矩阵中找出面积最大的全1矩阵.刚看到这道题会比较无从下手,b…

题目链接:http://acm.hust.edu.cn/vjudge/contest/126708#problem/J 题意:求一段子的连续最大和,只要每个数都大于0 那么就会一直增加,所以只要和0 比较就行,如果加上一数小于0了那么肯定要重新开始找,否则就不断更新最大值就行 AC代码: #include<stdio.h> #include<string.h> ]; int main() { int t,a,i,max,n,sum,start,ends,f; scanf("…

Description Given a sequence a[1],a[2],a[3]......a[n], your job is to calculate the max sum of a sub-sequence. For example, given (6,-1,5,4,-7), the max sum in this sequence is 6 + (-1) + 5 + 4 = 14. Input The first line of the input contains an inte…

A - Max Sum Plus Plus Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u Submit Status Practice HDU 1024 Appoint description: Description Now I think you have got an AC in Ignatius.L's "Max Sum" problem. To be a bra…

Max Sum Plus Plus Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 25639    Accepted Submission(s): 8884 Problem Description Now I think you have got an AC in Ignatius.L's "Max Sum" problem.…

测试样例之间输出空行,if(t>0) cout<<endl; 这样出最后一组测试样例之外,其它么每组测试样例之后都会输出一个空行. dp[i]表示以a[i]结尾的最大值,则:dp[i]=max(dp[i]+a[i],a[i]) 解释: 以a[i]结尾的最大值,要么是以a[i-1]为结尾的最大值+a[i],要么是a[i]自己本身,就是说,要么是连同之前的 构成一个多项的字串,要么自己单独作为一个字串,不会有其他的可能了. 状态规划的对状态的要求是:当前状态只与之前的状态有关,而且不影响下一…

A. Max Sum Plus Plus Now I think you have got an AC in Ignatius.L's "Max Sum" problem. To be a brave ACMer, we always challenge ourselves to more difficult problems. Now you are faced with a more difficult problem. Given a consecutive number seq…

Max Sum Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Problem Description Given a sequence a[1],a[2],a[3]......a[n], your job is to calculate the max sum of a sub-sequence. For example, given (6,-1,5,4,-7), the m…

LeetCode:Path Sum Given a binary tree and a sum, determine if the tree has a root-to-leaf path such that adding up all the values along the path equals the given sum. For example:Given the below binary tree and sum = 22, 5 / \ 4 8 / / \ 11 13 4 / \ \…

Max Sum Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 211310    Accepted Submission(s): 49611 Problem Description Given a sequence a[1],a[2],a[3]......a[n], your job is to calculate the max su…

http://acm.hdu.edu.cn/showproblem.php?pid=1024 Max Sum Plus Plus Problem Description Now I think you have got an AC in Ignatius.L's "Max Sum" problem. To be a brave ACMer, we always challenge ourselves to more difficult problems. Now you are fac…

题目大意:求一串数字中,几个连续数字加起来最大值,并确定起始和最末的位置. 思路:这是一题DP题,但是可以用尺取法来做.我一开始不会,也是看了某大神的代码,然后有人告诉我这是尺取法,现在会了. //尺取法 #include<stdio.h> #include<string.h> ]; main() { int t,flag; scanf("%d",&t); flag=t; while(t--) { memset(que,,sizeof(que)); ;…

#include <stdio.h> int main(){ int i,t,j,n,x; int start,end,temp,max,sum; scanf("%d",&t); ;i<t;i++){ temp=; max=-; sum=; scanf("%d",&n); ;j<n;j++){ scanf("%d",&x); sum+=x; if(sum>=max){ max=sum; sta…

HDU 1024 题目大意:给定m和n以及n个数,求n个数的m个连续子系列的最大值,要求子序列不想交. 解题思路:<1>动态规划,定义状态dp[i][j]表示序列前j个数的i段子序列的值,其中第i个子序列包括a[j], 则max(dp[m][k]),m<=k<=n 即为所求的结果 <2>初始状态: dp[i][0] = 0, dp[0][j] = 0; <3>状态转移: 决策:a[j]自己成为一个子段,还是接在前面一个子段的后面 方程: a[j]直接接在前面…

HDU 1003    相关链接   HDU 1231题解 题目大意:给定序列个数n及n个数,求该序列的最大连续子序列的和,要求输出最大连续子序列的和以及子序列的首位位置 解题思路:经典DP,可以定义dp[i]表示以a[i]为结尾的子序列的和的最大值,因而最大连续子序列及为dp数组中的最大值.   状态转移方程:dp[1] = a[1]; //以a[1]为结尾的子序列只有a[1]:  i >= 2时, dp[i] = max( dp[i-1]+a[i],  a[i] ); dp[i-1]+a[i…

虽然这道题看起来和 HDU 1024  Max Sum Plus Plus 看起来很像,可是感觉这道题比1024要简单一些 前面WA了几次,因为我开始把dp[22][maxn]写成dp[maxn][22]了,Orz 看来数组越界不一定会导致程序崩溃,也有可能返回一个错误的结果 dp[i][j]表示前j个数构成前i段所得到的最大值 状态转移方程: dp[i][j] = max{dp[i][j-1],  dp[i-1][j-len[i]] + sum[j] - sum[j-len[i]]} 分别对应…

事实上这连续发表的三篇是一模一样的思路,我就厚颜无耻的再发一篇吧! 题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1003 ---------------------------------------------------------------------------------------------------------------------------------------------------------- 欢迎光临天资小屋:…

Max Sum Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 154155    Accepted Submission(s): 35958 Problem Description Given a sequence a[1],a[2],a[3]......a[n], your job is to calculate the max su…

Problem Description Given a sequence a[1],a[2],a[3]......a[n], your job is to calculate the max sum of a sub-sequence. For example, given (6,-1,5,4,-7), the max sum in this sequence is 6 + (-1) + 5 + 4 = 14.   Input The first line of the input contai…