题目链接:戳我 题意:将一个长度为n的序列分为k段,使得总价值最大.一段区间的价值表示为区间内不同数字的个数 \(n<=35000,k<=50\) 开始想的转移方程是这个样子的--\(dp[i][j]\)表示前i个,分成j组,最大收益 然后转移方程为\(dp[i][j]=max(dp[i-1][j]+cur,dp[i-1][j-1]+1)\),其中cur表示这个数是否在当前组中出现过,判断可以用set来搞. 但是--不对!!!! 原因是同一种最大收益可能有不同的分组方式,而不同的分组方式显然具…

B. Bakery time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Masha wants to open her own bakery and bake muffins in one of the n cities numbered from 1 to n. There are m bidirectional roads,…

Bakery 题目链接: http://codeforces.com/contest/707/problem/B Description Masha wants to open her own bakery and bake muffins in one of the n cities numbered from 1 to n. There are m bidirectional roads, each of whose connects some pair of cities. To bake…

面包店烹制面包及蛋糕,由n个销售员卖出.当有顾客进店购买面包或蛋糕时,应先在取号机上取号,然后等待叫号,若有销售员空闲时便叫下一号,试用信号量和PV操作写出Bakery算法的同步程序. 设计要求 1)为每个销售员/顾客产生一个进程/线程,设计正确的同步算法: 2)每个顾客进入面包店后,即时显示"Entered",还同时显示共有几名顾客: 3)至少有个顾客: 4)多个销售员/顾客程序须共享操作函数代码. 专业程序代写c++程序代写…

D. The Bakery time limit per test:2.5 seconds memory limit per test:256 megabytes input:standard input output:standard output Some time ago Slastyona the Sweetmaid decided to open her own bakery! She bought required ingredients and a wonder-oven whic…

Some time ago Slastyona the Sweetmaid decided to open her own bakery! She bought required ingredients and a wonder-oven which can bake several types of cakes, and opened the bakery. Soon the expenses started to overcome the income, so Slastyona decid…

B. The Bakery time limit per test 2.5 seconds memory limit per test 256 megabytes input standard input output standard output Some time ago Slastyona the Sweetmaid decided to open her own bakery! She bought required ingredients and a wonder-oven whic…

B. Bakery 题目连接: http://www.codeforces.com/contest/707/problem/B Description Masha wants to open her own bakery and bake muffins in one of the n cities numbered from 1 to n. There are m bidirectional roads, each of whose connects some pair of cities.…

834D - The Bakery 思路:dp[i][j]表示到第j个数为止分成i段的最大总和值. dp[i][j]=max{dp[i-1][x]+c(x+1,j)(i-1≤x≤j-1)},c(x+1,j)表示x+1到j的不同的值. 用线段树维护一下最大值. 上图最后一个点取不到,不解释,不明白请评论. 代码: #include<bits/stdc++.h> using namespace std; #define ll long long #define ls rt<<1,l,m…

CF833B The Bakery LG传送门 线段树优化DP. 其实这是很久以前就应该做了的一道题,由于颓废一直咕在那里,其实还是挺不错的一道题. 先考虑\(O(n^2k)\)做法:设\(f[i][j]\)表示\(1\)到\(i\)之间分割\(j\)次得到的最大值,\(g[i][j]\)表示\(i\)到\(j\)之间不同的颜色个数. 转移就是: \[f[i][j]=max \{f[t][j-1]+g[t+1][i] \} \qquad t \in [1,i)\] 但是时空都无法承受,考虑优化.…