[ACM] POJ 1068 Parencodings(模拟)
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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:
#include <iostream>#include <string.h>#include <algorithm>using namespace std;int p[20],w[20];bool vis[40];//注意范围,题目中n<=20是n对括号,不是单个括号的个数.int main(){ int t;cin>>t; int n; while(t--) { string s; cin>>n; for(int i=1;i<=n;i++) cin>>p[i]; p[0]=0; for(int i=1;i<=n;i++)//构造s串 { for(int j=1;j<=(p[i]-p[i-1]);j++) s+="("; s+=")"; } int k=1; memset(vis,0,sizeof(vis)); for(int i=0;i<2*n;i++) { int cnt=1; if(s[i]==')')//遇到右括号 { for(int j=i-1;j>=0;j--)//回溯 { if(s[j]==')') cnt++; if(s[j]=='('&&!vis[j])//和小括号匹配 { vis[j]=1; break; } } w[k++]=cnt; } } for(int i=1;i<=n;i++) cout<<w[i]<<" "; cout<<endl; } return 0;}
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