
动态规划
文章平均质量分 82
madaidao
这个作者很懒,什么都没留下…
展开
-
Codeforces Round #728 (Div. 1) B. Tree Array(期望、dp)
题目链接:https://codeforces.com/contest/1540/problem/B题目大意: n个节点的树。初始的时候,等概率随机选择一个点标记,接来随机选择一个与标记点相连的未标记点来标记,直到所有的点都被标记。根据点被标记的顺序,生成一个数列。这个数列的逆序对的期望个数为多少个?题解: 我们单独计算每对逆序对出现的概率,也就是对于,a在b之前被标记的概率。 我们可以分别计算出每个点第一个选时,逆序对出现的概率。 ...原创 2021-06-26 21:10:58 · 345 阅读 · 0 评论 -
Deltix Round, Spring 2021 D. Love-Hate(随机+动态规划 or 搜索)
题目链接:https://codeforces.com/contest/1523/problem/D题目大意:原创 2021-06-01 19:45:12 · 254 阅读 · 1 评论 -
Codeforces Round #723 (Div. 2) C2. Potions (Hard Version) (贪心、dp、数据结构)
题目链接:https://codeforces.com/contest/1526/problem/C2题目大意:从左到右有杯酒,每杯酒有一个健康值,喝掉一杯酒后,人的健康值会加。初始的时候人的健康值为0,在最左边的酒,从左到右往后喝,对于一杯酒可以选择喝和不喝,人的健康值在整个过程不能小于0,求最多可以喝几杯酒?题解:方法1:dp表示前杯酒喝了杯最大的健康值这种方法复杂度个人的解法:贪心若当前酒健康值为-4且能喝,后面有杯健康值为-1,那么显然后面这杯健康值为-1的酒是.原创 2021-05-31 18:05:03 · 253 阅读 · 1 评论 -
Educational Codeforces Round 109 (Rated for Div. 2) D. Armchairs (dp)
题目链接:https://codeforces.com/contest/1525/problem/D题目大意:个数的0,1序列,a[i]=0或1。1的个数一定不超过n的一半。我们需要把所有的1和某个0配对,一个0只能和一个1配对,若和配对,则会消耗的资源。求所有1完成配对后,总资源消耗最少是多少?题解:个人的解法:首先我们可以想一个朴素的贪心匹配方法:我们从前往后扫描,若当前扫描到的数为1,则与目前扫描过的未匹配过的最近的0匹配;若目前扫描到0,则与目前扫描过的未匹配过的1匹配。我们可以原创 2021-05-20 16:25:45 · 258 阅读 · 0 评论 -
LeetCode 2021 春季赛组队赛 解题报告
LCP 33. 蓄水题目链接:https://leetcode-cn.com/problems/o8SXZn/题解:暴力枚举蓄水操作的次数计算最优解即可代码如下:class Solution {public: int storeWater(vector<int>& bucket, vector<int>& vat) { int sum=0; for(int i=0;i<int(vat.size());i原创 2021-05-09 17:48:50 · 332 阅读 · 0 评论 -
leetcode春季赛 LCP 31. 变换的迷宫 (动态规划)
题目链接解题思路首先我们容易想到一个简单的动态规划方法:用dp[t][x][y][s][fx][fy]表示在时间t时,s(0或1)表示临时消除法是否使用,以及永久消除法已经用在点(fx,fy),点(x,y)能否到达。用递推来计算出所有dp状态的结果。但是,这个方法的时间复杂度位O(Tnm2nm5),其中5为转移复杂度,最坏情况下约为10^9,复杂度太高。我们来考虑如何优化这个dp表达式,显然时间t,坐标(x,y),状态s都无法优化。我们把优化重点放到永久消除卡上。分类讨论:1,如果我们使用永原创 2021-04-06 18:00:45 · 258 阅读 · 0 评论 -
CodeForces 1389G Directing Edges 边联通分量缩点+树形dp+rerooting technique
题目链接:https://codeforces.com/contest/1389/problem/GG. Directing Edgestime limit per test4 secondsmemory limit per test512 megabytesinputstandard inputoutputstandard outputYou are given an undirected connected graph consisting ofnnverti..原创 2020-10-12 19:32:33 · 414 阅读 · 0 评论 -
poj 1742 Coins(dp)
题目链接CoinsTime Limit: 3000MS Memory Limit: 30000KTotal Submissions: 29601 Accepted: 10031DescriptionPeople in Silverland use coins.They have coins of value原创 2014-12-28 17:16:50 · 483 阅读 · 0 评论 -
poj 2288 Islands and Bridges(状压dp)
题目链接Islands and BridgesTime Limit: 4000MS Memory Limit: 65536KTotal Submissions: 9199 Accepted: 2386DescriptionGiven a map of islands and bridges that con原创 2014-12-30 14:53:31 · 457 阅读 · 0 评论 -
codeforces 505C Mr. Kitayuta, the Treasure Hunter(dp)
题目链接C. Mr. Kitayuta, the Treasure Huntertime limit per test1 secondmemory limit per test256 megabytesinputstandard inputoutputstandard outputThe Shuseki I原创 2015-01-19 22:05:35 · 902 阅读 · 0 评论 -
poj 1651 Multiplication Puzzle(区间dp)
题目链接Multiplication PuzzleTime Limit: 1000MS Memory Limit: 65536KTotal Submissions: 6612 Accepted: 4029DescriptionThe multiplication puzzle is played with a ro原创 2014-12-30 19:05:05 · 567 阅读 · 0 评论 -
poj 3661 Running(dp)
题目链接RunningTime Limit: 1000MS Memory Limit: 65536KTotal Submissions: 5279 Accepted: 1966DescriptionThe cows are trying to become better athletes, so Bessie原创 2014-12-30 19:27:09 · 425 阅读 · 0 评论 -
poj 2955 Brackets(区间dp)
题目链接BracketsTime Limit: 1000MS Memory Limit: 65536KTotal Submissions: 3470 Accepted: 1792DescriptionWe give the following inductive definition of a “regular原创 2015-01-05 21:46:45 · 363 阅读 · 0 评论 -
poj 1141 Brackets Sequence(区间dp)
题目链接Brackets SequenceTime Limit: 1000MS Memory Limit: 65536KTotal Submissions: 26254 Accepted: 7394 Special JudgeDescriptionLet us define a regular bracke原创 2015-01-05 21:21:02 · 357 阅读 · 0 评论 -
poj 3280 Cheapest Palindrome(区间dp)
题目链接Cheapest PalindromeTime Limit: 2000MS Memory Limit: 65536KTotal Submissions: 6117 Accepted: 2981DescriptionKeeping track of all the cows can be a tricky原创 2015-01-05 20:52:27 · 392 阅读 · 0 评论 -
zoj 3847 Collect Chars(dp)
题目链接Collect CharsTime Limit: 2 Seconds Memory Limit: 65536 KB Bob was playing MC and was punished by Alice. Bob was trapped in a maze and there were some characters on some specif原创 2015-01-11 21:30:00 · 745 阅读 · 0 评论 -
zoj 3845 Fixed Point(数位dp)
题目链接Fixed PointTime Limit: 2 Seconds Memory Limit: 65536 KB In mathematics, a fixed point (sometimes shortened to fixpoint, also known as an invariant point) of a function is an e原创 2015-01-11 21:18:24 · 762 阅读 · 0 评论 -
Hdu 2089 不要62(数位dp)
题目链接不要62Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 21020 Accepted Submission(s): 7180Problem Description杭州人称那些傻乎乎粘嗒嗒的人为6原创 2014-12-01 22:55:10 · 350 阅读 · 0 评论 -
Hdu 3853 LOOPS(概率dp)
题目链接LOOPSTime Limit: 15000/5000 MS (Java/Others) Memory Limit: 125536/65536 K (Java/Others)Total Submission(s): 2726 Accepted Submission(s): 1126Problem DescriptionAkemi Homura i原创 2015-01-07 16:55:53 · 483 阅读 · 0 评论 -
Hdu 2476 String painter(区间dp)
题目链接String painterTime Limit: 5000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1853 Accepted Submission(s): 820Problem DescriptionThere ar原创 2015-01-06 15:44:33 · 508 阅读 · 0 评论 -
poj 3071 Football(概率dp)
题目链接FootballTime Limit: 1000MS Memory Limit: 65536KTotal Submissions: 3284 Accepted: 1679DescriptionConsider a single-elimination football tournament involv原创 2015-01-07 17:10:48 · 372 阅读 · 0 评论 -
light oj 1422(区间dp)
题意:一个人先后要参加n个舞会,已知他n个舞会要穿什么样的衣服。一件衣服他脱了以后,就不会再穿。他可以穿了一种衣服以后,在外面套另一种衣服,再在外面套另一种衣服........。问他最少要准备多少件衣服?题解:原问题是参加完区间[1,n]的舞会,最少要准备多少件衣服。如果第1个舞会要穿的衣服和第i个舞会要穿的衣服相同,假设我们的策略是让第1个舞会和第i个舞会穿同一件衣服,那么区间[2,i-1]原创 2015-01-06 14:46:18 · 357 阅读 · 0 评论 -
Codeforces 149D Coloring Brackets(区间dp)
题目链接Coloring Bracketstime limit per test2 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard outputOnce Petya read a problem about a bracket原创 2015-01-06 17:11:01 · 395 阅读 · 0 评论 -
zoj 3469 Food Delivery(区间dp)
题目链接Food DeliveryTime Limit: 2 Seconds Memory Limit: 65536 KB When we are focusing on solving problems, we usually prefer to stay in front of computers rather than go out for lunc原创 2015-01-06 11:20:16 · 393 阅读 · 0 评论 -
Uvalive 6434 - Number Assignment(dp)
题目链接Uvalive 6434 - Number Assignment题意:有N个数原创 2014-08-01 15:38:26 · 1334 阅读 · 0 评论 -
Hdu 4507 吉哥系列故事——恨7不成妻(求平方和的数位dp)
题目 链接吉哥系列故事——恨7不成妻Time Limit: 1000/500 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others)Total Submission(s): 1752 Accepted Submission(s): 505Problem Description 单身!原创 2014-12-04 10:48:54 · 427 阅读 · 0 评论 -
zoj 2432 Greatest Common Increasing Subsequence(最长公共上升子序列)
题目链接Greatest Common Increasing SubsequenceTime Limit: 2 Seconds Memory Limit: 65536 KB Special Judge You are given two sequences of integer numbers. Write a program to determin原创 2015-01-07 20:09:17 · 478 阅读 · 0 评论 -
Hdu 5188 zhx and contest(暴搜+剪枝or暴力枚举or背包)
题目链接zhx and contestTime Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 546 Accepted Submission(s): 192Problem DescriptionAs原创 2015-03-25 12:00:40 · 654 阅读 · 0 评论 -
topcoder SRM 654 DIV2 1000 SuccessiveSubtraction2 题解(dp)
题意:有一个公式: a1-a2-a3-a4-a5......-anq次修改,每次修改一个位置上的数。 输出每次修改完以后,对于这个式子最多加两个括号后的最大值n题解:每次修改完以后,用动态规划的方法,O(n)的复杂度求最优解。用dp[i][j][k],表示当前在第i位,还有j个括号没有用,前面有k个有效的前括号的最优值。代码如下:// BE原创 2015-03-30 20:35:01 · 434 阅读 · 0 评论 -
Hdu 5262 最强密码(dp)
题目链接最强密码Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 78 Accepted Submission(s): 42Problem Description由于近来密码库被盗的现象层出不穷,度度熊原创 2015-06-09 19:24:40 · 969 阅读 · 0 评论 -
codeforces 152E Garden(状态压缩dp)
题目链接E. Gardentime limit per test2 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard outputVasya has a very beautiful country gard原创 2015-05-25 19:21:21 · 850 阅读 · 0 评论 -
Hdu 5236 Article(dp)
题目链接ArticleTime Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 404 Accepted Submission(s): 119Problem DescriptionAs the term is g原创 2015-06-12 15:45:24 · 600 阅读 · 0 评论 -
poj 1160 Post Office(四边形不等式优化dp)
转载自:http://www.cnblogs.com/staginner/archive/2012/03/12/2391925.html POJ_1160 我们可以用f[i][j]表示建好i个邮局时覆盖到第j个村庄的最优解,那么就可以得到f[i][j]=min{f[i-1][k]+w[k+1][j]}(k 这个题目还可以用四边形不等式去优化,实际上转载 2015-06-12 18:41:39 · 662 阅读 · 0 评论 -
Hdu 4734 F(x) (数位dp)
题目链接F(x)Time Limit: 1000/500 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2272 Accepted Submission(s): 861Problem DescriptionFor a decimal numbe原创 2015-06-12 20:39:48 · 351 阅读 · 0 评论 -
poj 1734 Sightseeing trip(floyd求最小环并输出方案)
题目链接Sightseeing tripTime Limit: 1000MS Memory Limit: 65536KTotal Submissions: 5736 Accepted: 2220 Special JudgeDescriptionThere is a travel agency in Adel原创 2015-11-22 21:43:22 · 446 阅读 · 0 评论 -
codeforces 601C(概率dp)
题目链接C. Kleofáš and the n-thlontime limit per test1 secondmemory limit per test256 megabytesinputstandard inputoutputstandard outputKleofáš is participating in原创 2015-11-27 21:36:42 · 947 阅读 · 0 评论 -
CodeForce 1153D(树上动态规划+贪心)
题目链接题意:N个节点的树,有K个叶子节点,需要将1,2,3...K,K个数任意填入K个叶子节点,每个叶子节点填一个数,每个数只能填入一个叶子节点。每个节点上有个一个运算,运算有两种Max和Min。若运算为Max,则该节点的值等于所有儿子节点的最大值;若运算为Min,该节点的值等于所有儿子节点的最小值。问,根节点可能出现的最大值?题解:关于树的问题,可以考虑树上动态规划的方法。 ...原创 2019-04-29 16:36:04 · 248 阅读 · 0 评论 -
大视野在线测评 1010 [HNOI2008]玩具装箱toy (动态规划)
题目链接1010: [HNOI2008]玩具装箱toyTime Limit: 1 Sec Memory Limit: 162 MBSubmit: 7027 Solved: 2670[Submit][Status][Discuss]DescriptionP教授要去看奥运,但是他舍不下他的玩具,于是他决定把所有的玩具运到北京。他使用自己的压缩器进行压缩,其可原创 2015-06-05 19:43:39 · 999 阅读 · 0 评论 -
Hdu 1059 Dividing(dp)
题目链接DividingTime Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 19596 Accepted Submission(s): 5514Problem DescriptionMarsha and B原创 2015-06-04 16:34:52 · 502 阅读 · 0 评论 -
poj 2923 Relocation(状压dp)
题目链接RelocationTime Limit: 1000MS Memory Limit: 65536KTotal Submissions: 2577 Accepted: 1052DescriptionEmma and Eric are moving to their new house they bough原创 2015-05-27 19:29:37 · 386 阅读 · 0 评论