POJ 1068 Parencodings(模拟)
2012-02-27 20:20
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Parencodings
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
Source
Tehran 2001
解题报告:这道题的题意就是给我们P序列,让我们求S序列,P-sequence表示第i个‘)’之前有几个‘(’,而S-sequence是表示在第i个“)”和它相匹配的(号中有几对相匹配的“()”的个数,包括它本身;现根据P序列求出括号的排列顺序,再求出S序列即可,主要就是找规律!
代码如下:
Time Limit: 1000MS | Memory Limit: 10000K | |
Total Submissions: 13696 | Accepted: 8145 |
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
Source
Tehran 2001
解题报告:这道题的题意就是给我们P序列,让我们求S序列,P-sequence表示第i个‘)’之前有几个‘(’,而S-sequence是表示在第i个“)”和它相匹配的(号中有几对相匹配的“()”的个数,包括它本身;现根据P序列求出括号的排列顺序,再求出S序列即可,主要就是找规律!
代码如下:
#include <iostream> #include <cstdio> #include <cstring> const int N = 50; int pse , wse , fuhao ; int visit ; int main() { int t, n, i, j, k; scanf("%d", &t); while (t --) { memset(pse, 0, sizeof(pse)); memset(wse, 0, sizeof(wse)); memset(fuhao, 0, sizeof(fuhao)); memset(visit, 0, sizeof(visit)); scanf("%d", &n); for (i = 1; i <= n; ++i) { scanf("%d", &pse[i]); } int count; for (i = 1, count = 1; i <= n; ++i) { int num = pse[i] - pse[i -1];//两个相邻的有括号中间的左括号的个数 while (num --) { fuhao[count ++] = 0;//符号为左括号时 } fuhao[count ++] = 1; } k = 1; for (i = 1; i <= 2 * n; ++i) { if (fuhao[i]) { count = 1; for (j = i - 1; j >= 1; -- j) { if (visit[j]==0 && !fuhao[j])//找到与之相匹配的(的位置 { visit[j] = 1; break; } else if (visit[j] && !fuhao[j])//求中间的左括号(的个数 { count ++; } } wse[k ++] = count; } } for (i = 1; i < k - 1; ++i) { printf("%d ", wse[i]); } printf("%d\n", wse[i]); } return 0; }
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