2015 ACM/ICPC Asia Regional Changchun Online 题意:n个池塘,删掉度数小于2的池塘,输出池塘数为奇数的连通块的池塘容量之和. 思路:两个dfs模拟就行了 #include <iostream> #include <cstdio> #include <fstream> #include <algorithm> #include <cmath> #include <deque> #include…

Time Limit: 1500/1000 MS (Java/Others)     Memory Limit: 131072/131072 K (Java/Others) Problem Description Betty owns a lot of ponds, some of them are connected with other ponds by pipes, and there will not be more than one pipe between two ponds. Ea…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5438 Ponds Time Limit: 1500/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 2237    Accepted Submission(s): 707 Problem Description Betty owns a lot of ponds, som…

题意:给定一个图,然后让你把边数为1的结点删除,然后求连通块结点数为奇的权值和. 析:这个题要注意,如果删除一些结点后,又形成了新的边数为1的结点,也应该要删除,这是坑,其他的,先用并查集判一下环,然后再找连通环. 代码如下: #include <cstdio> #include <string> #include <cstdlib> #include <cmath> #include <iostream> #include <cstrin…

Ponds Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://acm.hdu.edu.cn/contests/contest_showproblem.php?pid=1001&cid=621 Description Betty owns a lot of ponds, some of them are connected with other ponds by pipes, and there will not be more than one…

Ponds Time Limit: 1500/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 282    Accepted Submission(s): 86 Problem Description Betty owns a lot of ponds, some of them are connected with other ponds by pipes, and…

题目: 给出一个无向图,将图中度数小于等于1的点删掉,并删掉与他相连的点,直到不能在删为止,然后判断图中的各个连通分量,如果这个连通分量里边的点的个数是奇数,就把这些点的权值求和. 思路: 先用拓扑排序删点并更新各个点的度数,然后用并查集判断各个连通分量里边的点个数的奇偶性就ok了. 代码: #include <bits/stdc++.h> #include <cstdio> #include <cstring> #include <iostream> #i…

目录 题目地址 题干 代码和解释 参考 题目地址 hdu5438 题干 代码和解释 解答本题时参考了一篇代码较短的博客,比较有意思,使用了STL vector二维数组. 可以结合下面的示例代码理解: #include<iostream> #include<vector> using namespace std; int main() { vector<int> n[100]; int i; for(i=0;i<100;i++){ n[i].clear(); } n…

描述 我们可以把由“0”和“1”组成的字符串分为三类:全“0”串称为B串,全“1”串称为I串,既含“0”又含“1”的串则称为F串. FBI树是一种二叉树1,它的结点类型也包括F结点,B结点和I结点三种.由一个长度为2^N的“01”串S可以构造出一棵FBI树T,递归的构造方法如下: 1) T的根结点为R,其类型与串S的类型相同: 2) 若串S的长度大于1,将串S从中间分开,分为等长的左右子串S1和S2:由左子串S1构造R的左子树T1,由右子串S2构造R的右子树T2. 现在给定一个长度为2^N的“0…

HDU.5692 Snacks ( DFS序 线段树维护最大值 ) 题意分析 给出一颗树,节点标号为0-n,每个节点有一定权值,并且规定0号为根节点.有两种操作:操作一为询问,给出一个节点x,求从0号节点开始到x节点,所能经过的路径的权值最大为多少:操作二为修改,给出一个节点x和值val,将x的权值改为val. 可以看出是树上修改问题.考虑的解题方式有DFS序+线段树,树链剖分,CXTree.由于后两种目前还不会,选择用DFS序来解决. 首先对树求DFS序,在求解过程当中,顺便求解树上前缀和(p…

Ponds Time Limit: 1500/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 3234    Accepted Submission(s): 997 Problem Description Betty owns a lot of ponds, some of them are connected with other ponds by pipes, a…

嗯... 题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1035 这道题比较简单,但自己一直被卡,原因就是在读入mp这张字符图的时候用了scanf被卡... 注意初始化和dfs边界:如果超出图或者曾经被标记过则输出 AC代码: #include<cstdio> #include<cstring> #include<iostream> using namespace std; ][]; ][]; inline void df…

题意 不断删去度数为1的点,最后求有奇数个点的联通块的权值之和. 分析 存边的时候,要头尾都存这个边.用dfs或者队列删点,再用并查集或者dfs确定联通块,然后统计联通块的点数,最后累加. 我自己写的超时,然后参考了网上的题解.真郁闷. 代码 并查集 #include<cstdio> #include<queue> #include<vector> #define ll long long using namespace std; const int N = 1e4 +…

Problem Description There was no donkey ,) , the down-right cell ,N-) and the cell below the up-left cell ,)..... A × grid is shown below: The donkey lived happily until it saw a tiger far away. The donkey had never seen a tiger ,and the tiger had ne…

Robot Motion Problem Description A robot has been programmed to follow the instructions in its path. Instructions for the next direction the robot is to move are laid down in a grid. The possible instructions are  N north (up the page) S south (down…

题意:       给你一些按键顺序,让你输出每一步中概率最大的那个单词,这里的概率计算方 法好好看看别弄错了,一开始就是因为弄错了,各种wa,比如 abc 1 ,ab 1,那么 ab 的概率就是2 ,而不是4,一开始我误认为是所有单词累加后再把每个单词的每个字母累加作为当前单词的概率的,结果各种wa. 思路:       先建一颗字典树,为了是节省内存,方便更新,和快速查询,其实hash也可以 ,不过我自己一般都是用map去hasn,目测这个题目map去hash会TLE,因为要设计到拆串和各种…

题目链接 http://acm.hdu.edu.cn/search.php?action=listproblem Problem Description Ladies and gentlemen, please sit up straight.Don't tilt your head. I'm serious.For n given strings S1,S2,⋯,Sn, labelled from 1 to n, you should find the largest i (1≤i≤n) su…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5047 解题报告:问一个“M”型可以把一个矩形的平面最多分割成多少块. 输入是有n个“M",现在已经推出这个公式应该是8 * n^2 - 7 * n + 1,但是这个n的范围达到了10^12次方,只要平方一次就超出long long  的范围了,怎么办呢,用大数? 都试过了,很奇怪,会超时,按照估算的话感觉不会,可能是中间结果比较大吧,这个还在思考,但是10^12平方一次乘以八也只达到了10^25次方…

Ponds Time Limit: 1500/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 533    Accepted Submission(s): 175 Problem Description Betty owns a lot of ponds, some of them are connected with other ponds by pipes, an…

Necklace/center> 题目连接: http://acm.hdu.edu.cn/showproblem.php?pid=5727 Description SJX has 2*N magic gems. N of them have Yin energy inside while others have Yang energy. SJX wants to make a necklace with these magic gems for his beloved BHB. To avoid…

题意: 给一个字符串,表示一颗树,要求你把它整理出来,节点从1开始编号,还要输出树边. 解法: 模拟即可.因为由括号,所以可以递归地求,用map存对应关系,np存ind->name的映射,每进入一层括号,使father = now, 遇到右括号')',则father = fa[father],用vector存每个节点的子节点,然后最后dfs输出即可. 代码: #include <iostream> #include <cstdio> #include <cstring&…

Cannon Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://acm.hdu.edu.cn/showproblem.php?pid=4499 Description In Chinese Chess, there is one kind of powerful chessmen called Cannon. It can move horizontally or vertically along the chess grid. At eac…

扫雷 Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 998    Accepted Submission(s): 289 Problem Description 扫雷游戏是晨晨和小璐特别喜欢的智力游戏,她俩最近沉迷其中无法自拔.该游戏的界面是一个矩阵,矩阵中有些格子中有一个地雷,其余格子中没有地雷. 游戏中,格子可能处于己知和未知的状态…

Car Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 25    Accepted Submission(s): 12 Problem Description Ruins is driving a car to participating in a programming contest. As on a very tight sche…

Fraction Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 0    Accepted Submission(s): 0 Problem Description Mr. Frog recently studied how to add two fractions up, and he came up with an evil ide…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1175 解题思路:从出发点开始DFS.出发点与终点中间只能通过0相连,或者直接相连,判断能否找出这样的路径. #include<cmath> #include<cstdio> #include<cstring> #include<iostream> #include<algorithm> using namespace std; #define N 1…

pid=4831" style="font-weight:normal">题目链接:hdu 4831 Scenic Popularity 题目大意:略. 解题思路:对于休闲区g[i][0]和g[i][1]记录的是近期的两个景点的id(仅仅有一个近期的话g[i][1]为0),对于景点来说.g[i][0]为-1(表示该id相应的是景点),g[i][1]为该景点的热度值.主要就是模拟,注意一些细节就能够了. #include <cstdio> #include &…

题目网址:http://acm.hdu.edu.cn/showproblem.php?pid=5547 题目: Sudoku Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)Total Submission(s): 2372    Accepted Submission(s): 804 Problem Description   Yi Sima was one of the be…

题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=5538 Problem Description Have you ever played the video game Minecraft? This game has been one of the world's most popular game in recent years. The world of Minecraft is made up of lots of 1×1×1 blocks…

题目链接: F - Auxiliary Set HDU - 5927 学习网址:https://blog.csdn.net/yiqzq/article/details/81952369题目大意一棵节点数为n的有根数,根节点为1,一开始所有的点都是重点,接下来有q次询问,每次询问把m个点变为轻点,问你树中还有多少个重点. 重点应该满足的条件为: 1.它本身是重点. 2.它为两个重点的最近公共祖先. 每次询问之后在下次询问前,所有的点都恢复为重点. 具体思路:对于每个点保存他的深度.因为每次输入的数…