标签归档:题解

Openjudge 鸣人和佐助

鸣人和佐助

总时间限制: 1000ms 内存限制: 65536kB
描述
佐助被大蛇丸诱骗走了,鸣人在多少时间内能追上他呢?

已知一张地图(以二维矩阵的形式表示)以及佐助和鸣人的位置。地图上的每个位置都可以走到,只不过有些位置上有大蛇丸的手下,需要先打败大蛇丸的手下才能到这些位置。鸣人有一定数量的查克拉,每一个单位的查克拉可以打败一个大蛇丸的手下。假设鸣人可以往上下左右四个方向移动,每移动一个距离需要花费1个单位时间,打败大蛇丸的手下不需要时间。如果鸣人查克拉消耗完了,则只可以走到没有大蛇丸手下的位置,不可以再移动到有大蛇丸手下的位置。佐助在此期间不移动,大蛇丸的手下也不移动。请问,鸣人要追上佐助最少需要花费多少时间?
输入
输入的第一行包含三个整数:M,N,T。代表M行N列的地图和鸣人初始的查克拉数量T。0 < M,N < 200,0 ≤ T < 10 后面是M行N列的地图,其中@代表鸣人,+代表佐助。*代表通路,#代表大蛇丸的手下。 输出 输出包含一个整数R,代表鸣人追上佐助最少需要花费的时间。如果鸣人无法追上佐助,则输出-1。 样例输入 样例输入1 4 4 1 #@## **## ###+ **** 样例输入2 4 4 2 #@## **## ###+ **** 样例输出 样例输出1 6 样例输出2 4 继续阅读

Openjudge Charm Bracelet

Charm Bracelet

总时间限制:
1000ms
内存限制:
65536kB
描述

Bessie has gone to the mall’s jewelry store and spies a charm bracelet. Of course, she’d like to fill it with the best charms possible from the N(1 ≤ N≤ 3,402) available charms. Each charmiin the supplied list has a weight Wi(1 ≤ Wi≤ 400), a ‘desirability’ factor Di(1 ≤ Di≤ 100), and can be used at most once. Bessie can only support a charm bracelet whose weight is no more than M(1 ≤ M≤ 12,880).

Given that weight limit as a constraint and a list of the charms with their weights and desirability rating, deduce the maximum possible sum of ratings.

输入
Line 1: Two space-separated integers: N and M
Lines 2..N+1: Line i+1 describes charm i with two space-separated integers: Wi and Di
输出
Line 1: A single integer that is the greatest sum of charm desirabilities that can be achieved given the weight constraints
样例输入
4 6
1 4
2 6
3 12
2 7
样例输出
23
来源
USACO 2007 December Silver

继续阅读

Openjudge 棋盘问题

棋盘问题

总时间限制:
1000ms
内存限制:
65536kB
描述
在一个给定形状的棋盘(形状可能是不规则的)上面摆放棋子,棋子没有区别。要求摆放时任意的两个棋子不能放在棋盘中的同一行或者同一列,请编程求解对于给定形状和大小的棋盘,摆放k个棋子的所有可行的摆放方案C。
输入
输入含有多组测试数据。
每组数据的第一行是两个正整数,n k,用一个空格隔开,表示了将在一个n*n的矩阵内描述棋盘,以及摆放棋子的数目。 n <= 8 , k <= n
当为-1 -1时表示输入结束。
随后的n行描述了棋盘的形状:每行有n个字符,其中 # 表示棋盘区域, . 表示空白区域(数据保证不出现多余的空白行或者空白列)。
输出
对于每一组数据,给出一行输出,输出摆放的方案数目C (数据保证C<2^31)。
样例输入
2 1
#.
.#
4 4
...#
..#.
.#..
#...
-1 -1
样例输出
2
1
来源
蔡错@pku

继续阅读

Openjudge A Knight’s Journey

A Knight’s Journey

总时间限制:
1000ms
内存限制:
65536kB
描述
Background
The knight is getting bored of seeing the same black and white squares again and again and has decided to make a journey
around the world. Whenever a knight moves, it is two squares in one direction and one square perpendicular to this. The world of a knight is the chessboard he is living on. Our knight lives on a chessboard that has a smaller area than a regular 8 * 8 board, but it is still rectangular. Can you help this adventurous knight to make travel plans?Problem
Find a path such that the knight visits every square once. The knight can start and end on any square of the board.
输入
The input begins with a positive integer n in the first line. The following lines contain n test cases. Each test case consists of a single line with two positive integers p and q, such that 1 <= p * q <= 26. This represents a p * q chessboard, where p describes how many different square numbers 1, . . . , p exist, q describes how many different square letters exist. These are the first q letters of the Latin alphabet: A, . . .
输出
The output for every scenario begins with a line containing “Scenario #i:”, where i is the number of the scenario starting at 1. Then print a single line containing the lexicographically first path that visits all squares of the chessboard with knight moves followed by an empty line. The path should be given on a single line by concatenating the names of the visited squares. Each square name consists of a capital letter followed by a number.
If no such path exist, you should output impossible on a single line.
样例输入
3
1 1
2 3
4 3
样例输出
Scenario #1:
A1

Scenario #2:
impossible

Scenario #3:
A1B3C1A2B4C2A3B1C3A4B2C4
来源
TUD Programming Contest 2005, Darmstadt, Germany

继续阅读

Openjudge 红与黑

红与黑

总时间限制:
1000ms
内存限制:
65536kB
描述
有一间长方形的房子,地上铺了红色、黑色两种颜色的正方形瓷砖。你站在其中一块黑色的瓷砖上,只能向相邻的黑色瓷砖移动。请写一个程序,计算你总共能够到达多少块黑色的瓷砖。
输入
包括多个数据集合。每个数据集合的第一行是两个整数W和H,分别表示x方向和y方向瓷砖的数量。W和H都不超过20。在接下来的H行中,每行包括W个字符。每个字符表示一块瓷砖的颜色,规则如下
1)‘.’:黑色的瓷砖;
2)‘#’:白色的瓷砖;
3)‘@’:黑色的瓷砖,并且你站在这块瓷砖上。该字符在每个数据集合中唯一出现一次。
当在一行中读入的是两个零时,表示输入结束。
输出
对每个数据集合,分别输出一行,显示你从初始位置出发能到达的瓷砖数(记数时包括初始位置的瓷砖)。
样例输入
6 9 
....#. 
.....# 
...... 
...... 
...... 
...... 
...... 
#@...# 
.#..#. 
0 0
样例输出
45
来源
1979

继续阅读

数据结构 邓俊辉 PA#2 旅行商(TSP) 题解

旅行商(TSP)


Description

Shrek is a postman working in the mountain, whose routine work is sending mail to n villages. Unfortunately, road between villages is out of repair for long time, such that some road is one-way road. There are even some villages that can’t be reached from any other village. In such a case, we only hope as many villages can receive mails as possible.

Shrek hopes to choose a village A as starting point (He will be air-dropped to this location), then pass by as many villages as possible. Finally, Shrek will arrived at village B. In the travelling process, each villages is only passed by once. You should help Shrek to design the travel route.

Input

There are 2 integers, n and m, in first line. Stand for number of village and number of road respectively.

In the following m line, m road is given by identity of villages on two terminals. From v1 to v2. The identity of village is in range [1, n].

Output

Output maximum number of villages Shrek can pass by.

Example

Input

4 3
1 4
2 4
4 3

Output

3

Restrictions

1 <= n <= 1,000,000

0 <= m <= 1,000,000

These is no loop road in the input.

Time: 2 sec

Memory: 256 MB

Hints

Topological sorting

描述

Shrek是一个大山里的邮递员,每天负责给所在地区的n个村庄派发信件。但杯具的是,由于道路狭窄,年久失修,村庄间的道路都只能单向通过,甚至有些村庄无法从任意一个村庄到达。这样我们只能希望尽可能多的村庄可以收到投递的信件。

Shrek希望知道如何选定一个村庄A作为起点(我们将他空投到该村庄),依次经过尽可能多的村庄,路途中的每个村庄都经过仅一次,最终到达终点村庄B,完成整个送信过程。这个任务交给你来完成。

输入

第一行包括两个整数n,m,分别表示村庄的个数以及可以通行的道路的数目。

以下共m行,每行用两个整数v1和v2表示一条道路,两个整数分别为道路连接的村庄号,道路的方向为从v1至v2,n个村庄编号为[1, n]。

输出

输出一个数字,表示符合条件的最长道路经过的村庄数。

样例

见英文题面

限制

1 ≤ n ≤ 1,000,000

0 ≤ m ≤ 1,000,000

输入保证道路之间没有形成环

时间:2 sec

空间:256 MB

提示

拓扑排序

继续阅读

数据结构 邓俊辉 PA#2 真二叉树重构(Proper Rebuild)题解

真二叉树重构(Proper Rebuild)


Description

In general, given the preorder traversal sequence and postorder traversal sequence of a binary tree, we cannot determine the binary tree.

Figure 1

In Figure 1 for example, although they are two different binary tree, their preorder traversal sequence and postorder traversal sequence are both of the same.

But for one proper binary tree, in which each internal node has two sons, we can uniquely determine it through its given preorder traversal sequence and postorder traversal sequence.

Label n nodes in one binary tree using the integers in [1, n], we would like to output the inorder traversal sequence of a binary tree through its preorder and postorder traversal sequence.

Input

The 1st line is an integer n, i.e., the number of nodes in one given binary tree,

The 2nd and 3rd lines are the given preorder and postorder traversal sequence respectively.

Output

The inorder traversal sequence of the given binary tree in one line.

Example

Input

5
1 2 4 5 3
4 5 2 3 1

Output

4 2 5 1 3

Restrictions

For 95% of the estimation, 1 <= n <= 1,000,00

For 100% of the estimation, 1 <= n <= 4,000,000

The input sequence is a permutation of {1,2…n}, corresponding to a legal binary tree.

Time: 2 sec

Memory: 256 MB

Hints

Figure 2

In Figure 2, observe the positions of the left and right children in preorder and postorder traversal sequence.

描述

一般来说,给定二叉树的先序遍历序列和后序遍历序列,并不能确定唯一确定该二叉树。

(图一)

比如图一中的两棵二叉树,虽然它们是不同二叉树,但是它们的先序、后序遍历序列都是相同的。

但是对于“真二叉树”(每个内部节点都有两个孩子的二叉树),给定它的先序、后序遍历序列足以完全确定它的结构。

将二叉树的n个节点用[1, n]内的整数进行编号,输入一棵真二叉树的先序、后序遍历序列,请输出它的中序遍历序列。

输入

第一行为一个整数n,即二叉树中节点的个数。

第二、三行为已知的先序、后序遍历序列。

输出

仅一行,给定真二叉树的中序遍历序列。

样例

见英文题面

限制

对于95%的测例:1 ≤ n ≤ 1,000,000

对于100%的测例:1 ≤ n ≤ 4,000,000

输入的序列是{1,2…n}的排列,且对应于一棵合法的真二叉树

时间:2 sec

空间:256 MB

提示

观察左、右孩子在先序、后序遍历序列中的位置

重温视频05e5-3

目测一辈子也忘不了二叉树重构怎么写了…………
继续阅读

PATest 2017春季 ZigZagging on a Tree (30) 题解

1127. ZigZagging on a Tree (30)

Suppose that all the keys in a binary tree are distinct positive integers. A unique binary tree can be determined by a given pair of postorder and inorder traversal sequences. And it is a simple standard routine to print the numbers in level-order. However, if you think the problem is too simple, then you are too naive. This time you are supposed to print the numbers in “zigzagging order” — that is, starting from the root, print the numbers level-by-level, alternating between left to right and right to left. For example, for the following tree you must output: 1 11 5 8 17 12 20 15.

Input Specification:

Each input file contains one test case. For each case, the first line gives a positive integer N (<= 30), the total number of nodes in the binary tree. The second line gives the inorder sequence and the third line gives the postorder sequence. All the numbers in a line are separated by a space.

Output Specification:

For each test case, print the zigzagging sequence of the tree in a line. All the numbers in a line must be separated by exactly one space, and there must be no extra space at the end of the line.

Sample Input:

8
12 11 20 17 1 15 8 5
12 20 17 11 15 8 5 1

Sample Output:

1 11 5 8 17 12 20 15

继续阅读

Openjudge 复杂的整数划分问题

复杂的整数划分问题

总时间限制:
200ms
内存限制:
65536kB
描述
将正整数n 表示成一系列正整数之和,n=n1+n2+…+nk, 其中n1>=n2>=…>=nk>=1 ,k>=1 。
正整数n 的这种表示称为正整数n 的划分。
输入
标准的输入包含若干组测试数据。每组测试数据是一行输入数据,包括两个整数N 和 K。
(0 < N <= 50, 0 < K <= N)
输出
对于每组测试数据,输出以下三行数据:
第一行: N划分成K个正整数之和的划分数目
第二行: N划分成若干个不同正整数之和的划分数目
第三行: N划分成若干个奇正整数之和的划分数目
样例输入
5 2
样例输出
2
3
3
提示
第一行: 4+1, 3+2,
第二行: 5,4+1,3+2
第三行: 5,1+1+3, 1+1+1+1+1+1

继续阅读

THU2017spring 1-3 Interview 题解

THU2017spring 1-3 Interview


描述

某公司在对应聘者做过一轮笔试之后,从中选出n 人继续进行面试,每位应聘者被分配了一个整数ID。

为公平起见,组织者决定利用会议室外的圆桌,按以下方法“随机”确定面试顺序:第一个到达的应聘者在圆桌周围任意选择一个位置坐下;此后到达的每位应聘者都从前一应聘者出发,沿逆时针方向围圆桌走过m 人(前一应聘者算作走过的第1 人,同一人可能经过多次),并紧邻第m 人右侧就座;所有应聘者到齐后,从最后到达者出发,绕圆桌以顺时针方向为序进行面试。

这里假定应聘者到达的时刻互异,且相对的就坐位置确定后,左、右两人之间总能插入一把椅子。

试编写一个程序,确定面试顺序。

输入

共2行。

第1行包含两个整数, n和m。

第2行包含n个整数,表示先后到达的n个应聘者的ID。

输出

共1行。以空格分隔的n个整数,分别表示顺次进行面试的应聘者的ID。

输入样例

5 3
6 7 8 9 10 

输出样例

10 6 8 9 7

限制

1 ≤ n ≤ 10^3

1 ≤ m ≤ 2*n

输入的ID保证在int类型的范围内。

时间:1 sec

空间:256 MB

提示

一级提示

● 循环链表

二级提示

● 因为找座位按逆时针方向,面试按顺时针发现,所以建议使用双向循环链表。

● 往链表中插入元素时,要小心地修改前驱指针和后继指针。

● 一次性new n个节点,比分n次每次new一个节点快一些,但本题的数据规模不至于一定要这么做。

继续阅读