poj 1068 -- Parencodings
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:
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 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。
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
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 }
Do one thing , and do it well !