POJ 1068 (13.10.11)
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Parencodings
Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 17713 Accepted: 10660
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
题意: 给出字符串的表达式P, 得到另一种表达式W
表达式P的规则是这样的: P1, P2, Pi...Pn, Pi表示的是, 第i个右括号左边有Pi个左括号
表达式W的规则是这样的: W1, W2, Wi...Wn, Wi表示的是, 第i个右括号与其对应的左括号形成的一对括号中, 有几对括号, 包括自己本身;
做法: 由P规则得出字符串, 再由该字符串从左至右逐个找右括号, 一对一对查, 查一对, 标记成"##", 以便下次查的时候不会重复, 以及可以得知这里已有一对括号, 便于统计
AC代码:
#include<stdio.h>int main() { int t; scanf("%d", &t); while(t--) { char str[50]; int ans[50]; int n, formal, now; scanf("%d", &n); n--; scanf("%d", &formal); int i, pos; for(pos = 0; pos < formal; pos++) str[pos] = '('; str[pos] = ')'; while(n--) { scanf("%d", &now); int num; num = now - formal + pos + 1; for(i = pos+1; i < num; i++) str[i] = '('; str[i] = ')'; pos = i; formal = now; } int count; int tpos = 0; for(i = 0 ; i <= pos; i++) { count = 0; if(str[i] == ')') { for(int j = i-1; j >= 0; j--) { if(str[j] == '#') count++; if(str[j] == '(') { str[j] = '#'; str[i] = '#'; break; } } ans[tpos++] = (count / 2) + 1; } } for(int i = 0; i < tpos; i++) { if(i != tpos-1) printf("%d ", ans[i]); else printf("%d\n", ans[i]); } } return 0;}
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