poj 1068 -- Parencodings

时间：2014-05-24 09:16:54 阅读：191 评论：0 收藏：0 [点我收藏+]

Parencodings

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 19068		Accepted: 11503

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:


	S		(((()()())))

	P-sequence	    4 5 6666

	W-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

2
6
4 5 6 6 6 6
9 
4 6 6 6 6 8 9 9 9

Sample Output

1 1 1 4 5 6
1 1 2 4 5 1 1 3 9

简单模拟题，给出若干左右括号，能写出两种数列，其一:在第i个右括号前有多少个左括号pi就是几。 其二：找距离第i个右括号最近的左括号（没有配过对的）。
此题给出其一队列，求其二，模拟即可，不多说。

 1 /*======================================================================
 2  *           Author :   kevin
 3  *         Filename :   Parencodings.cpp
 4  *       Creat time :   2014-05-23 13:41
 5  *      Description :
 6 ========================================================================*/
 7 #include <iostream>
 8 #include <algorithm>
 9 #include <cstdio>
10 #include <cstring>
11 #include <queue>
12 #include <cmath>
13 #define clr(a,b) memset(a,b,sizeof(a))
14 #define M 50
15 using namespace std;
16 char str[M];
17 int vis[M];
18 int main(int argc,char *argv[])
19 {
20     int t,n;
21     scanf("%d",&t);
22     while(t--){
23         scanf("%d",&n);
24         clr(str,0);
25         clr(vis,0);
26         int a,cc = 0,cnt = 0;
27         for(int i = 0; i < n; i++){
28             scanf("%d",&a);
29             for(int j = cc; j < a; j++){
30                 str[cnt++] = ‘(‘;
31             }
32             str[cnt++] = ‘)‘;
33             cc = a;
34         }
35         int kong = 0;
36         for(int i = 0; i < 2*n; i++){
37             if(str[i] == ‘)‘){
38                 int steps = 1;
39                 for(int j = i-1; j >= 0; j--){
40                     if(str[j] == ‘(‘ && !vis[j]){
41                         if(kong != n-1){
42                             printf("%d ",steps);
43                             kong++;
44                         }
45                         else{
46                             printf("%d\n",steps);
47                         }
48                         vis[j] = 1;
49                         break;
50                     }
51                     if(str[j] == ‘(‘ && vis[j]){
52                         steps++;
53                     }
54                 }
55             }
56         }
57     }
58     return 0;
59 }

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poj 1068 -- Parencodings

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原文地址：http://www.cnblogs.com/ubuntu-kevin/p/3747602.html

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