hdu1024 Max Sum Plus Plus的另一种解法】的更多相关文章

传送门:http://acm.hdu.edu.cn/showproblem.php?pid=1024 http://www.51nod.com/onlineJudge/questionCode.html#!problemId=1053 [题解] 原来的解法:http://www.cnblogs.com/galaxies/p/hdu1024.html 原来的解法思路是正着,我们加数,现在考虑删数 我们先特判几种情况(可以全选正,全选正还不够) 现在只剩下一种情况,段数<正数段数 我们可以把数字合并…
Max Sum Plus Plus Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 17164    Accepted Submission(s): 5651 Problem Description Now I think you have got an AC in Ignatius.L's "Max Sum" problem…
Max Sum Plus Plus Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 28136    Accepted Submission(s): 9810 Problem Description Now I think you have got an AC in Ignatius.L's "Max Sum" problem.…
Max Sum Plus Plus Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 41885    Accepted Submission(s): 15095 Problem Description Now I think you have got an AC in Ignatius.L's "Max Sum" problem…
题目链接:http://acm.split.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): 31798    Accepted Submission(s): 11278 Problem Description Now I think…
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 sequence S 1, S 2, S 3,…
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…
动态规划,给定长度为n(≤1e6)的整数数组和整数m,选取m个连续且两两无交集的子区间,求所有方案中使得区间和最大的最大值. dp[i][j]表示结束位置(最后一个区间最后一个元素的位置)为i且选取区间数为j的最大值. 容易得到以下状态转移方程:   又:         考虑到数组的规模和j的更新特征,使用一维滚动数组取代二维数组,最外层循环枚举j到m即可. 用dp[0][i]表示dp[i][j], dp[1][i]表示max(dp[k][j-1]) (k≤i). 复杂度为O(n^2) . #…
状态:d(i,j)它代表前j划分数i部并且包括第一j最佳结果时的数.g(i,j)表示前j划分数i最好的结果时,段,g(m,n)结果,需要. 本题数据较大.需採用滚动数组.注意:这题int类型就够用了,开long long可能会TLE. 用滚动数组后,g[j]表示分成i段时最优结果,最后求出的g[n]即为结果 #include<iostream> #include<cstdio> #include<cstring> #include<algorithm> #i…
链接:http://acm.hdu.edu.cn/showproblem.php?pid=1024 #include<iostream> #include<vector> #include<algorithm> #include<cmath> #include<cstring> #include<queue> #include<cstdio> #include<unordered_map> #define in…