POJ 1068：Parencodings（模拟）

Radar Installation

Time limit:1000 ms Memory limit:10000 kB OS:Linux

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Problem 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

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

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题意：

有一串括号，根据括号生成两个序列：
P：从左到右，每个右括号之前的左括号数量。
W：从左到右，每个右括号匹配左括号，生成括号对中共包含的括号对数量。
现给出P串，求W串。

解题思路：

根据P串很容易还原出括号串，第一个数是4，说明第一个右括号前有4个左括号，第二个数是5，说明到上一个右括号前有5-4个左括号，依次类推，可以还原出括号串。
接着就是用栈来匹配括号了，很容易发现，中间包含多少个括号对，可以直接用左括号和右括号的下标来计算，因此，碰到左括号时，把其下标压入栈，碰到右括号时，出栈，由两个下标计算出结果。

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Code：

#include <iostream>#include <algorithm>#include <cstdio>#include <cstring>#include <stack>using namespace std;int  a[25];stack<int> s;char str[100];int main(){    int T;    scanf("%d",&T);    while(T--)    {        int n;        scanf("%d",&n);        for(int i=0;i<n;i++)        {            scanf("%d",&a[i]);        }        int len=0,cnt=a[0];        for(int i=0;i<n;i++)        {            if(i==0)                cnt=a[0];            else                cnt=a[i]-a[i-1];            for(int i=0;i<cnt;i++)                str[len++]='(';            str[len++]=')';        }        //str[len]='\0';        //cout<<str<<endl;//此处已经还原出括号串        for(int i=0;i<len;i++)        {            if(str[i]=='(')                s.push(i);            else            {                printf("%d",(i-s.top()+1)/2);                if(i==len-1)                    printf("\n");                else                    printf(" ");                s.pop();            }        }    }    return 0;}