How Many Trees? Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 3952    Accepted Submission(s): 2238 Problem Description A binary search tree is a binary tree with root k such that any node v re…

How Many Trees? Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 3317    Accepted Submission(s): 1922 Problem Description A binary search tree is a binary tree with root k such that any node v r…

裸的卡特兰数 C++#include<iostream> #include<cstdio> using namespace std; #define base 10000 #define len 100 void multiply(int a[],int max,int b) { ; ;i>=;i--) { array+=b*a[i]; a[i]=array%base; array/=base; } } void divide(int a[],int max,int b) {…

题目 题意:给你一个数字n,问你将1~n这n个数字,可以组成多少棵不同的二叉搜索树. 1,2,5,14--根据输出中的规律可以看出这是一个卡特兰数的序列.于是代用卡特兰数中的一个递推式: 因为输入可取到100,用无符号位计算最高可计算33个卡特兰数,所以可以用java中的大数 import java.math.BigInteger; import java.util.Scanner; public class Main { public static void main(String[] arg…

[题目链接]http://acm.hdu.edu.cn/showproblem.php?pid=1134 [解题背景]这题不会做,自己推公式推了一段时间,将n=3和n=4的情况列出来了,只发现第n项与第n-1项有关系,上网搜索的时候发现是组合数学中关于Catalan(卡特兰)数的运用. 以下资料来自网络,整理以备记录学习: [来源链接] http://baike.baidu.com/view/2499752.htm http://blog.163.com/lz_666888/blog/stati…

题目:http://acm.hdu.edu.cn/showproblem.php?pid=1693 第一道插头 DP ! 直接用二进制数表示状态即可. #include<cstdio> #include<cstring> #include<algorithm> #define ll long long using namespace std; ,M=(<<)+; int n,m,bin[N];ll dp[N][N][M]; int b[N][N]; void…

How Many Trees? Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 3382    Accepted Submission(s): 1960 Problem Description A binary search tree is a binary tree with root k such that any node v re…

Surround the Trees Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 10803    Accepted Submission(s): 4187 Problem Description There are a lot of trees in an area. A peasant wants to buy a rope t…

Surround the Trees:http://acm.hdu.edu.cn/showproblem.php?pid=1392 题意: 在给定点中找到凸包,计算这个凸包的周长. 思路: 这道题找出凸包上的点后,s数组中就是按顺序的点,累加一下距离就是周长了. #include <algorithm> #include <iterator> #include <iostream> #include <cstring> #include <cstdli…

题目链接:HDU 1392 Problem Description There are a lot of trees in an area. A peasant wants to buy a rope to surround all these trees. So at first he must know the minimal required length of the rope. However, he does not know how to calculate it. Can you…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1392 这里介绍一种求凸包的算法:Graham.(相对于其它人的解释可能会有一些出入,但大体都属于这个算法的思想,同样可以解决凸包问题) 相对于包裹法的n*m时间,Graham算法在时间上有很大的提升,只要n*log(n)时间就够了.它的基本思想如下: 1.首先,把所有的点按照y最小优先,其次x小的优先排序 2.维护一个栈,用向量的叉积来判断新插入的点跟栈顶的点哪个在外围,如果栈顶的点在当前插入的点的…

插头DP基础题的样子...输入N,M<=11,以及N*M的01矩阵,0(1)表示有(无)障碍物.输出哈密顿回路(可以多回路)方案数... 看了个ppt,画了下图...感觉还是挺有效的... 参考http://www.cnblogs.com/kuangbin/archive/2012/10/02/2710343.html 以及推荐cd琦的论文ppthttp://wenku.baidu.com/view/4fe4ac659b6648d7c1c74633.html 向中学生学习~~ 感觉以后可能还会要…

题目链接 USACO 第6章,第一题是一个插头DP,无奈啊.从头看起,看了好久的陈丹琦的论文,表示木看懂... 大体知道思路之后,还是无法实现代码.. 此题是插头DP最最简单的一个,在一个n*m的棋盘上,有些点能走,有些点不能走,可以走一条回路,也可以多回路,把所有点走完,有多少种走法.. 这题的背景还是dota,还是屠夫,还是吃树...我还是不会玩屠夫啊... 学习此题,看的下面的大神的博客,图画很棒,位运算又学了一个新用法. http://blog.csdn.net/xymscau/arti…

第一道(可能也是最后一道)插头dp.... 总算是领略了它的魅力... #include<iostream> #include<cstdio> #include<cstring> #include<algorithm> using namespace std; ][],dp[][][(<<)+]; void work(long long x) { scanf("%I64d%I64d",&n,&m); ;i<…

题目地址:http://acm.hdu.edu.cn/showproblem.php?pid=1392 题意:给出一些点的坐标,求最小的凸多边形把所有点包围时此多边形的周长. 解法:凸包ConvexHull 模板题 #include<iostream> #include<cstdio> #include<cstring> #include<cmath> #include<cstdlib> #include<algorithm> #de…

 Count the Trees  Another common social inability is known as ACM (Abnormally Compulsive Meditation). This psychological disorder is somewhat common among programmers. It can be described as the temporary (although frequent) loss of the faculty of sp…

题目大意:要求你将全部非障碍格子都走一遍,形成回路(能够多回路),问有多少种方法 解题思路: 參考基于连通性状态压缩的动态规划问题 - 陈丹琦 下面为代码 #include<cstdio> #include<algorithm> #include<cstring> using namespace std; #define N 12 #define S (1 << 12) int n, m; long long dp[N][N][S]; int cas = 1…

There are a lot of trees in an area. A peasant wants to buy a rope to surround all these trees. So at first he must know the minimal required length of the rope. However, he does not know how to calculate it. Can you help him? The diameter and length…

Query on The Trees Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65768/65768 K (Java/Others)Total Submission(s): 4091    Accepted Submission(s): 1774 Problem Description We have met so many problems on the tree, so today we will have a que…

Surround the Trees Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Problem Description There are a lot of trees in an area. A peasant wants to buy a rope to surround all these trees. So at first he must know the mi…

http://acm.hdu.edu.cn/showproblem.php?pid=1392 题目大意: 二维平面给定n个点,用一条最短的绳子将所有的点都围在里面,求绳子的长度. 解题思路: 凸包的模板.凸包有很多的算法.这里用Adrew. 注意这几组测试数据 1 1 1 3 0 0 1 0 2 0 输出数据 0.00 2.00 #include<cmath> #include<cstdio> #include<algorithm> using namespace st…

Surround the Trees Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 6728    Accepted Submission(s): 2556 Problem Description There are a lot of trees in an area. A peasant wants to buy a rope to…

Surround the Trees Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 6812    Accepted Submission(s): 2594 Problem Description There are a lot of trees in an area. A peasant wants to buy a rope to…

Surround the Trees Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 7043    Accepted Submission(s): 2688 Problem Description There are a lot of trees in an area. A peasant wants to buy a rope to…

Problem Description Most of us know that in the game called DotA(Defense of the Ancient), Pudge is a strong hero in the first period of the game. When the game goes to end however, Pudge is not a strong hero any more.So Pudge’s teammates give him a n…

题意:给一个n*m的矩阵,为1时代表空格子,为0时代表障碍格子,问如果不经过障碍格子,可以画一至多个圆的话,有多少种方案?(n<12,m<12) 思路: 这题不需要用到最小表示法以及括号表示法. 以一个非障碍格子为单位进行DP转移,所以可以用滚动数组.只需要保存m+1个插头的状态,其中有一个是右插头,其他都是下插头,若有插头的存在,该位为1,否则为0,初始时都是0. 需要考虑的是,(1)如果两个边缘都是插头,那么必须接上它们:(2)如果仅有一边是插头,则延续插头,可以有两个延续的方向(下和右)…

[题目分析] 吃树. 直接插头DP,算是一道真正的入门题目. 0/1表示有没有插头 [代码] #include <cstdio> #include <cstring> #include <iostream> #include <algorithm> using namespace std; #define u64 unsigned long long #define F(i,j,k) for (int i=j;i<=k;++i) int n,m,t,a…

Problem Description There are a lot of trees in an area. A peasant wants to buy a rope to surround all these trees. So at first he must know the minimal required length of the rope. However, he does not know how to calculate it. Can you help him? The…

题目链接 题意: 给出一个n*m大小的01矩阵,在其中画线连成封闭图形,其中对每一个值为1的方格,线要恰好穿入穿出共两次,对每一个值为0的方格,所画线不能经过. 参考资料: <基于连通性状态压缩的动态规划问题> ——陈丹琦 2008年国家集训队论文 递推过程中,按照 遍历行->遍历行上每一格->遍历 “轮廓线跨过该格时所有可能的状态变化”   的顺序 这样复杂度是 O(n*m*2m+1)  (m+1是因为轮廓线上有m个单元是与列数对应的,另有一单独的竖线单元) 问题关键点在于解决 …

题目是说给出一个数字,然后以1到这个数为序号当做二叉树的结点,问总共有几种组成二叉树的方式.这个题就是用卡特兰数算出个数,然后因为有编号,不同的编号对应不同的方式,所以结果是卡特兰数乘这个数的阶乘种方案.因为数字比较大,所以要用高精度的方法也就是用字符数组来做,我分别写了三个函数,一个算加法,一个算乘法,最后一个打表,等打出表来最后只要判断一下输入的数是第几个,直接输出就行了,下面是我的代码,第一次写高精度的这种大数处理,可能看上去比较繁琐= = #include<iostream> #inc…