Parencodings(poj模拟法)

jinzk123

于 2016-10-12 15:28:04 发布

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分类专栏：算法POJ

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本文介绍了Parencodings问题，详细解析了如何从原括号串生成两种数字密码串：p序列和w序列。p序列通过从括号对的右括号往左数到最左边的左括号；w序列则记录括号对内所有右括号的数量。

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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

题意：
对于给出的原括号串，存在两种数字密码串：

1.p序列：当出现匹配括号对时，从该括号对的右括号开始往左数，直到最前面的左括号数，就是pi的值。

2.w序列：当出现匹配括号对时，包含在该括号对中的所有右括号数（包括该括号对）,就是wi的值。

简单模拟
用数组a[i]表示第i个右括号和第i+1个右括号间的左括号数
然后逐渐找和右括号匹配的左括号所处的位置
i-j

#include <iostream>
#include <cstdio>
using namespace std;

int main()
{
    int n,t,i,a[30],p[30],w[30],j;
    scanf("%d",&t);
    while(t--)
    {
       scanf("%d",&n);
       for(i=1;i<=n;i++)
        scanf("%d",&p[i]);
        a[0]=p[1];
       for(i=1;i<=n-1;i++)
       {
           a[i]=p[i+1]-p[i];
       }
       for(i=1;i<=n;i++)//一共n个右括号
       {
           for(j=i-1;j>=0;j--)//首先第i个到第i-1个右括号之间,若j--那右括号的数量就加1
           {
               if(a[j]>0)//相邻两右括号之间左括号的数量存在则可匹配(a[j]到a[j+1])
               {
                  a[j]--;
                  break;
               }
           }
           w[i]=i-j;//第i个右括号与第j个和第j+1个右括号之间的左括号匹配,i-j便是括号对之间右括号的数量
       }
       for(i=1;i<=n;i++)
       {
           if(i==n)
            printf("%d\n",w[i]);
           else printf("%d ",w[i]);
       }
    }
    return 0;
}