dp..dp(i, j)表示画两个点为i-j, i的最优答案. dp(i, j) = min{ dp(i-j, k) } + cost[i] (1≤k≤M-j) 令f(i, j) = min{dp(i, j)}, 那么转移时间下降为O(1).然后滚动数组..这道题卡空间..时间复杂度O(NM) -------------------------------------------------------------------------------- #include<cstdio> #in…

183. Painting the balls time limit per test: 0.25 sec.memory limit per test: 4096 KB input: standard inputoutput: standard output Petya puts the N white balls in a line and now he wants to paint some of them in black, so that at least two black balls…

题意:给n个白球,选其中一些涂为黑色,且给了涂第i个球的花费为ci,要求每m个连续的球中至少有两个黑球,问最小花费是多少? 容易想到一个方程dp[i][j]=min{dp[k][i]}+c[j] dp[i][j]表示最后上色的两个球分别是第i和第j个球(i<j)时的最优解. 我们从条件每m个连续的球中至少有两个黑球可以得出k的范围j-m<=k<i 如果我们直接按上面这个方程去做的话时间复杂度是O(n*m*m). #include <iostream> #include <…

时间限制:0.25s 空间限制:4M 题意:  在n(n<=10000)个球中,给若干个球涂色,每个球涂色的代价为Ci,使得任意连续m(m<=100)个球中有至少两个球被涂了色. Solution: 首先很直接地能想到一个DP的状态转移方程 f[i][j] 代表,当前涂第i个球,且前面最近一个被涂色的球为j的最小代价 f[i][j]=min(f[j][k])+Ci,   k>i+1-m 分析这个转移方程的时间复杂度是O(n*m*m)在此题的数据范围中高达10^8 显然我们需要更好的解法…

Description Petya puts the \(N\) white balls in a line and now he wants to paint some of them in black, so that at least two black balls could be found among any \(M\) successive balls. Petya knows that he needs \(C_i\) milliliters of dye exactly to…

http://acm.sgu.ru/problemset.php?contest=0&volume=1 101 Domino 欧拉路 102 Coprime 枚举/数学方法 103 Traffic Lights 最短路 104 Little Shop of Flowers 动态规划 105 Div 3 找规律 106 The Equation 扩展欧几里德 107 987654321 Problem 找规律 108 Self-numbers II 枚举+筛法递推 109 Magic of Dav…

-----------------------------最优化问题------------------------------------- ----------------------常规动态规划  SOJ1162 I-Keyboard  SOJ1685 Chopsticks SOJ1679 Gangsters SOJ2096 Maximum Submatrix  SOJ2111 littleken bg SOJ2142 Cow Exhibition  SOJ2505 The County…

之前做了一个基于python的tkinter的小球完全碰撞游戏: 今天利用业余时间,写了一个功能要强大一些的小球完全碰撞游戏: 游戏名称: 小球完全弹性碰撞游戏规则: 1.游戏初始化的时候,有5个不同颜色的小球进行碰撞 2.玩家可以通过在窗口中单击鼠标左键进行增加小球个数 3.玩家可以通过在窗口中单击鼠标右键进行删减小球个数 4.玩家可以通过键盘的方向键:上,右键进行对小球加速 5.玩家可以通过键盘的方向键:下,左键进行对小球减速 6.玩家可以按键盘:f键实现全屏显示 7.玩家可以按键盘:Esc…

121. Bridges painting time limit per test: 0.25 sec. memory limit per test: 4096 KB New Berland consists of N (1£ N£ 100) islands, some of them are connected by bridges. There can be no more than one bridge between any pair of islands. Mr. President…

题目大意:有个一无向图,给所有的边染色,如果一个点连接的边超过两个,那么最少要染一个白色和一个黑色,能否给整个图染色?不能输出“No solution”. 分析:引用连接 http://edward-mj.com/archives/445 首先构建dfs树,无向图dfs树具有的一大优点是该点只会向自己的祖先或子孙有非树边. 然后按深度交替染色.返祖边与自己的儿子涂同样的颜色. 如果dfs树中根结点度超过1,那么就找一条边染不同的颜色. 否则看根结点是否满足条件,如果不是,那么找一个与根结点相连的…

原题地址 题意: 新百慕大由N个岛屿组成,在岛屿之间有一些连接它们的桥.在任意两个岛屿之间,最多只有一座桥连接它们.总统先生下达命令,要求给所有桥上色. 每一座桥能被染成 白色 或者 黑色. 每一个岛屿至少有一座白色的桥和一座黑色的桥(当然,如果只有一座桥就不存在这些问题) Solution: 很简单的dfs,水水就能过. 代码 #include <iostream> #include <cstring> #define INF 111 using namespace std; i…

Ball Painting 题目连接: http://codeforces.com/gym/100015/attachments Description There are 2N white balls on a table in two rows, making a nice 2-by-N rectangle. Jon has a big paint bucket full of black paint. (Don't ask why.) He wants to paint all the b…

C. Kyoya and Colored Balls Time Limit: 2000ms Memory Limit: 262144KB 64-bit integer IO format: %I64d      Java class name: (Any) Submit Status Kyoya Ootori has a bag with n colored balls that are colored with k different colors. The colors are labele…

今天讨论的是称球问题. No.3 13 balls problem You are given 13 balls. The odd ball may be either heavier or lighter. Find out the odd ball in 3 weightings. 分析与解答: 看到这道题,就想起来高中时候数学老师的一句话:“真正难的题不是那些很长的题,而是那些就几句话的题!!!”现在想想真是良训啊.又想到很多老师的话,感觉到失去方显弥足珍贵的名言,不禁唏嘘啊…… 有人认为…

Codeforces Round #256 (Div. 2) C C. Painting Fence time limit per test 1 second memory limit per test 512 megabytes input standard input output standard output Bizon the Champion isn't just attentive, he also is very hardworking. Bizon the Champion d…

水概率....SGU里难得的水题.... 495. Kids and Prizes Time limit per test: 0.5 second(s)Memory limit: 262144 kilobytes input: standardoutput: standard ICPC (International Cardboard Producing Company) is in the business of producing cardboard boxes. Recently the…

C16H:Magical Balls 总时间限制:  1000ms 内存限制:  262144kB 描述 Wenwen has a magical ball. When put on an infinite plane, it will keep duplicating itself forever. Initially, Wenwen puts the ball on the location (x0, y0) of the plane. Then the ball starts to dup…

Chrome Canary(Chrome “金丝雀版本”)目前已经支持Continuous painting mode,用于分析页面性能.这篇文章将会介绍怎么才能页面在绘制过程中找到问题和怎么利用这个新的工具来解决页面性能上的瓶颈. PS:最新版本的Chrome已经支持该功能查看页面的渲染时间我们采用Things We Left On The Moon by Dan Cederholm的页面来作为我们的例子页面.打开Chrome的Web Inspector(即按F12),选择Timeline页卡…

Vasya works as a watchman in the gallery. Unfortunately, one of the most expensive paintings was stolen while he was on duty. He doesn't want to be fired, so he has to quickly restore the painting. He remembers some facts about it. The painting is a…

sgu 101 - Domino Time Limit:250MS     Memory Limit:4096KB     64bit IO Format:%I64d & %I64u Description Dominoes – game played with small, rectangular blocks of wood or other material, each identified by a number of dots, or pips, on its face. The bl…

http://acm.sgu.ru/problem.php?contest=0&problem=495 题意:N个箱子M个人,初始N个箱子都有一个礼物,M个人依次等概率取一个箱子,如果有礼物则拿出礼物放回盒子,如果没有礼物则不操作.问M个人拿出礼物个数的期望.(N,M<=100000) #include <cstdio> using namespace std; double mpow(double a, int n) { double r=1; while(n) { if(n&…

http://acm.hdu.edu.cn/showproblem.php?pid=4710 Balls Rearrangement Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 735    Accepted Submission(s): 305 Problem Description Bob has N balls and A b…

题目链接:http://acm.sgu.ru/problem.php?contest=0&problem=455 Due to the slow 'mod' and 'div' operations with int64 type, all Delphi solutions for the problem 455 (Sequence analysis) run much slower than the same code written in C++ or Java. We do not gua…

319. Kalevich Strikes Back Time limit per test: 0.5 second(s)Memory limit: 65536 kilobytes input: standardoutput: standard And yet again the Berland community can see that talent is always multi-sided. The talent is always seeking new ways of self-ex…

题目大意 某人在打字机上打一个字符串,给出了他打每个字符出错的概率 q[i]. 打一个字符需要单位1的时间,删除一个字符也需要单位1的时间.在任意时刻,他可以花 t 的时间检查整个打出来的字符串,并且从当前光标删除到第一个出错的位置(留下的字符串要么为空,要么每个字符都是对的).问你,他正确的打完该字符串最少需要花费的时间的期望值是多少 字符串的长度小于等于3000 题意有点绕,具体可以看看原题是怎么描述的 做法分析 首先确定这是一个概率DP的问题. 定义状态 f[i] 表示正确的打完前 i 个…

104. Little shop of flowers time limit per test: 0.25 sec. memory limit per test: 4096 KB 问题: 你想要将你的花窗安排得最具美感.有F束花,每一束花都不一样,至少有F个按顺序排成一行的花瓶.花瓶从左到右,依次编号1-V.而花放置的位置是可以改变的,花依次编号1到F.花的序号有一个特征,即是序号决定了花束出现在花瓶里的顺序.例如,有两束花编号i和j,满足i<j,那么i所在的花瓶一定要在j所在花瓶的左边,即是i…

Dragon Balls Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2909    Accepted Submission(s): 1125 Problem Description Five hundred years later, the number of dragon balls will increase unexpecte…

Labeling Balls Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 9641 Accepted: 2636 Description Windy has N balls of distinct weights from 1 unit to N units. Now he tries to label them with 1 to N in such a way that: No two balls share the…

题目链接: Wall Painting Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2681    Accepted Submission(s): 857 Problem Description Ms.Fang loves painting very much. She paints GFW(Great Funny Wall) ev…

令一个点的属性值为:去除这个点以及与这个点相连的所有边后得到的连通分量的节点数的最大值. 则树的重心定义为:一个点,这个点的属性值在所有点中是最小的. SGU 134 即要找出所有的重心,并且找出重心的属性值. 考虑用树形DP. dp[u]表示割去u点,得到的连通分支的节点数的最大值. tot[u]记录以u为根的这棵子树的节点数总和(包括根). 则用一次dfs即可预处理出这两个数组.再枚举每个点,每个点的属性值其实为max(dp[u],n-tot[u]),因为有可能最大的连通分支在u的父亲及以上…