题目链接 Problem Description 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 th…

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…

题目地址:http://poj.org/problem?id=1050 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 a…

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…

一.动态规划算法 众所周知,递归算法时间复杂度很高为(2^n),而动态规划算法也能够解决此类问题,动态规划的算法的时间复杂度为(n^2).动态规划算法是以空间置换时间的解决方式,一开始理解起来可能比较困难,自己画画也许明白了很多. 二.动态规划算法分析 先举个例子: {7,0,0,0,0},{3,8,0,0,0},{8,1,0,0,0},{2,7,4,4,0},{4,5,2,6,5} 这个二维数组,求一下,顶层到底层,只能通过两端来相加的最大值(也就是说这棵树的最长路径). 分析: (1)第一步…

package findMax; /** * 连续子数组的最大和 * @author root * */ public class FindMax { static int[] data = {1,-2,3,10,-4,7,2,-5}; public static void main(String[] args) { // TODO Auto-generated method stub find(); } //分析数据规律 public static void find(){ int max =…

题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=1081 http://acm.xmu.edu.cn/JudgeOnline/problem.php?id=1031 题目大意: 给一个n*n(n<=100)的矩阵,求一个矩形覆盖的值最大是多少. 题目思路: [动态规划] 二维的最大字段和.先考虑一维的情况.f[i]=max(f[i-1]+a[i],a[i]) 只要之前的部分和大于零则一起取一定比只取当前位置的要优. 因此只要判断局部段的和是否大于零…

http://acm.hdu.edu.cn/showproblem.php?pid=1003 给出一个包含n个数字的序列{a1,a2,..,ai,..,an},-1000<=ai<=1000 求最大连续子段和及其起始位置和终止位置,很基础的动态规划(DP)问题,看完DP第一次做的DP题目 DP真的是一种很优美的算法,或者说思想,但是比较难理解,我对DP的理解还很浅薄 # include <stdio.h> # define INF 1000000000 int main() { i…

Max Sum Plus Plus Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 29942    Accepted Submission(s): 10516 Problem Description Now I think you have got an AC in Ignatius.L's "Max Sum" problem…

Max Sum Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 250714    Accepted Submission(s): 59365 Problem Description Given a sequence a[1],a[2],a[3]......a[n], your job is to calculate the max su…

HDU 1024 Max Sum Plus Plus (动态规划) 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 faced with a more difficult problem. Given…

Max Sum Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 193355 Accepted Submission(s): 45045 Problem Description Given a sequence a[1],a[2],a[3]--a[n], your job is to calculate the max sum of a su…

To the Max Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Problem 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 locate…

原创 http://acm.hdu.edu.cn/showproblem.php?pid=1003 题目要求求出一个序列里面的最大序列和,序列要求是连续的,给出最大序列和,序列首元素下标和尾元素下标,按特定的格式输出. 解题思路: 动态规划,我们可以将所有序列按以序列中的元素a[i](i=1~n)结尾进行分类,比如: 以a[1]结尾的序列有:a[1] 以a[2]结尾的序列有:a[1]a[2],a[2] 以a[3]结尾的序列有:a[1]a[2]a[3],a[2][3],a[3] ......  这…

题目链接:https://cn.vjudge.net/problem/HDU-1024 题意 给n, m和一个序列,找m个不重叠子串,使这几个子串内元素和的和最大. n<=1e6 例:1 3 1 2 3 答:6 (唯一的子串1 2 3) 思路 先顺便记录一下动态规划的一般解题思路: 原问题->子问题->状态->转移->边界 再顺便记录一下最大值最小化这类问题套路解法: 二分 贪心 不能二分的问题,贪心八九不离十. 一般是AB和BA这两个元素的顺序,不影响前后变化时,直接算目标…

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…

Max Sum Plus Plus Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 44371    Accepted Submission(s): 16084 Problem Description Now I think you have got an AC in Ignatius.L's "Max Sum" problem…

Max Sum Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 330535    Accepted Submission(s): 78678 Problem Description Given a sequence a[1],a[2],a[3]......a[n], your job is to calculate the max su…

解题思路: 本题在给定的集合中找到最大的子集合[子集合:集合的元素的总和,是所有子集合中的最大解.] 结果输出: 最大的子集合的所有元素的和,子集合在集合中的范围区间. 依次对元素相加,存到一个 sum 中,同时ans=sum;定义左右边界 left,right;临时左边界ll=1; 如果sum>ans,则ans=sum; 左边界 left=tem; right=i+1; 如果sum<0,则sum=0; tem=i+2; Ac code: #include<bits/stdc++.h&g…

题目地址:http://acm.hdu.edu.cn/showproblem.php?pid=1024 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 faced with a mor…

一般形式: $max\{min(ax+by+c,dF(x)+eG(y)+f)\},其中F(x)和G(y)是单调函数.$ 或 $min\{max(ax+by+c,dF(x)+eG(y)+f)\},其中F(x)和G(y)是单调函数.$ (以下用第一种形式讨论) (1)dF(x)随ax的增大而增大,eG(y)随by的增大而增大. ax和by均取最大值. (2)dF(x)随ax的增大而增大,eG(y)随by的增大而减小. ax一定取最大值,ax和dF(x)变成常数. 此时变成: $H(y)=max\{m…

http://acm.hdu.edu.cn/showproblem.php?pid=1024 Max Sum Plus Plus Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 21363    Accepted Submission(s): 7144 Problem Description Now I think you have go…

转载于acm之家http://www.acmerblog.com/hdu-1003-Max-Sum-1258.html Max Sum Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 242353    Accepted Submission(s): 57218 Problem Description Given a sequence…

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…

题目来源:http://acm.hdu.edu.cn/showproblem.php?pid=1024 (http://www.fjutacm.com/Problem.jsp?pid=1375) 题意:长度为n的序列里,m段不相关区间的最大和 思路:我们先要确定一个东西,就是状态,这里我用dp[i][j]表示前j个数在取a[j]情况下分i段的最大和: 那么我们为了找规律,可以先来一发Excel,就以样例为例子: 然后我们可以发现其实红圈里的8是状态dp[2][6](i=2, j=6),那么我们可…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1024 题意就是有n个数分成m段,求最大的和: dp[i][j]表示把 j 个数分成 i 段,选择第 j 个数的结果,而并不是当前的最优解, 那么要考虑的是第i个数的数是自己成一段还是和前面的成一段 所以dp[i][j]=max(dp[i][j-1]+a[j], Max+a[j]); 其中Max为前 j-1 个数字分成 i-1 段中的最大值: 由于题中n为100w,m不知道是多少,所以开二维数组可能不…

参考:https://www.cnblogs.com/yexiaozi/p/5749338.html #include <iostream> #include <cstdio> #include <cstring> using namespace std; ; int a[N],dp[N]; int n; void test() { cout<<"\n---------\n"; ;i<=n;i++) { cout<<'[…

Max Sum Plus Plus Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 22262    Accepted Submission(s): 7484   Problem Description Now I think you have got an AC in Ignatius.L's "Max Sum" problem. T…

Max Sum Plus PlusTime Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 37418    Accepted Submission(s): 13363 Problem DescriptionNow I think you have got an AC in Ignatius.L's "Max Sum" problem.…

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 faced with a more difficult problem. Given a consecutive number sequ…