poj 1068 Parencodings（模拟题）

题目链接：http://poj.org/problem?id=1068

Parencodings

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 25772 Accepted: 15159

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 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

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题意：给出每个右括号前面有几个左括号，求每个右括号包含几个括号

解析：将左括号记为1，右括号记为2，存到一个数组里面，遍历数组时每遇到右括号i，然后往前找第一个没标记的左括号j，这个范围内包含了（i - j + 1） / 2

代码：

#include<iostream>#include<cstdio>#include<algorithm>#include<cstring>#include<queue>#include<cmath>#include<map>#include<vector>#include<stack>#define N 29using namespace std;const int INF = 0x3f3f3f3f;int a[N], b[N<<1], used[N<<1];int main(){    int t, n;    scanf("%d", &t);    while(t--)    {        scanf("%d", &n);        for(int i = 1; i <= n; i++) scanf("%d", &a[i]);        int cnt = 0, cur = 0;        memset(used, 0, sizeof(used));        for(int i = 1; i <= n; i++)        {            while((cur++) < a[i]) b[cnt++] = 1;            b[cnt++] = 2;            cur = a[i];        }        int f = 0;        for(int i = 0; i < cnt; i++)        {            if(!used[i] && b[i] == 2)            {                for(int j = i - 1; j >= 0; j--)                {                    if(!used[j] && b[j] == 1)                    {                        used[j] = 1;                        a[f++] = (i - j + 1) / 2;                        break;                    }                }            }        }        for(int i = 0; i < n; i++)           printf("%d%c", a[i], i == n - 1 ? '\n' : ' ');    }    return 0;}