[ACM] POJ 1068 Parencodings(模拟)
来源:互联网 发布:股票自动交易 源码 编辑:程序博客网 时间:2024/05/29 08:31
Parencodings
Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 19352 Accepted: 11675
Description
Let S = s1 s2...s2n be a well-formed string of parentheses. S can be encoded in two different ways:
q By an integer sequence P = p1 p2...pn where pi is the number of left parentheses before the ith right parenthesis in S (P-sequence).
q By an integer sequence W = w1 w2...wn where for each right parenthesis, say a in S, we associate an integer which is the number of right parentheses counting from the matched left parenthesis of a up to a. (W-sequence).
Following is an example of the above encodings:
Write a program to convert P-sequence of a well-formed string to the W-sequence of the same string.
q By an integer sequence P = p1 p2...pn where pi is the number of left parentheses before the ith right parenthesis in S (P-sequence).
q By an integer sequence W = w1 w2...wn where for each right parenthesis, say a in S, we associate an integer which is the number of right parentheses counting from the matched left parenthesis of a up to a. (W-sequence).
Following is an example of the above encodings:
S(((()()())))P-sequence 4 5 6666W-sequence 1 1 1456
Write a program to convert P-sequence of a well-formed string to the W-sequence of the same string.
Input
The first line of the input contains a single integer t (1 <= t <= 10), the number of test cases, followed by the input data for each test case. The first line of each test case is an integer n (1 <= n <= 20), and the second line is the P-sequence of a well-formed string. It contains n positive integers, separated with blanks, representing the P-sequence.
Output
The output file consists of exactly t lines corresponding to test cases. For each test case, the output line should contain n integers describing the W-sequence of the string corresponding to its given P-sequence.
Sample Input
264 5 6 6 6 69 4 6 6 6 6 8 9 9 9
Sample Output
1 1 1 4 5 61 1 2 4 5 1 1 3 9
Source
Tehran 2001
题意为给一个只包含括号的字符串加密有两种方法:
方法一:用p数组表示,p[i]为第i个右括号左边一共有多少左括号
方法二:用w数组表示,w[i]表示当第i个括号左右匹配时,一共包括多少右括号
要求给定加密后的p数组,求出w数组。
可以根据给的p数组先求出字符串s, p[i]-p[i-1]为第i个右括号紧跟在它前面的有多少个左括号,求出s
遍历s,每次找到右括号,然后回溯,遇到右括号就计数(回溯前找到的那个也算),直到遇到与它匹配的左括号(vis[]=0),因为一个右括号有唯一的左括号匹配,所以一旦找到它的左括号,就用vis[]=1标记下。
代码:
0 0
- [ACM] POJ 1068 Parencodings(模拟)
- [ACM] POJ 1068 Parencodings(模拟)
- poj 1068 Parencodings(模拟)
- poj 1068 Parencodings(模拟)
- poj 1068 Parencodings(模拟)
- poj 1068 Parencodings(模拟)
- POJ 1068 Parencodings (模拟)
- poj 1068 Parencodings(模拟)
- poj 1068 Parencodings(模拟)
- POJ 1068 Parencodings(模拟)
- POJ 1068 Parencodings(模拟)
- POJ 1068:Parencodings(模拟)
- POJ 1068 Parencodings 模拟
- poj 1068 Parencodings (模拟)
- poj 1068 Parencodings 模拟
- POJ 1068-Parencodings(模拟)
- poj 1068 Parencodings(模拟)
- poj 1068 Parencodings 模拟
- 感知机介绍及实现
- [leetcode]89. Gray Code
- 深度学习准备之安装双系统
- jsp中无法正确显示<freamset></freamset>标签
- leetcode_middle_16_94. Binary Tree Inorder Traversal
- [ACM] POJ 1068 Parencodings(模拟)
- 沉浸状态栏的实现
- Solr简介
- 神经网络和深度学习简史(第一部分):从感知机到BP算法
- mysql常用错误分析与解决方法
- 1025
- 对一个二维数组中的数据排序,方法如下: 将整个数组中值最小的元素所在行调整为数组第一行, 将除第一行外的行中最小元素所在行调整为第2行, 将除第1,2行外的行中最小值元素所在行调整为第3行,以此类推
- MyBatis 源码分析——介绍
- 枚举