《ACM国际大学生程序设计竞赛题解I》——6.8

Poj1068：

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

  题目大意：这里有三个数组S、P、W，S是长度为2n是符合数学规律的括号字符串，P、W是依据S生成的长度为n的整数列。

  P：{p1,p2,p3…}第i个右括号左边的左括号数目，记为pi.

  W:{w1,w2,w3…}第i个右括号与其匹配的左括号，截取的那个区间段左括号的数目。

  现在给出P序列，求出对应的W序列。

  数理分析：很容易想到，首先我们应该根据P得到S，然后根据S和W的定义得到W。因此整个模拟流程分为如下的两个步骤：

  (一)P -> S:

  遍历整数列P，其含义是第i个右括号左边有的左括号数目,我们先将n个右括号画出，则第i个右括号和第i-1个右括号中间隔了P[i] – P[i-1]个左括号，按照这个规律，我们遍历P[i]，先生成P[i]-P[i-1]个左括号(P[0] = 0)，然后生成一个右括号，便可构造出S。

  (二)S –> W：

  我们遍历S，设置数组left[i]记录第i个左括号在字符串数组S中的下标，则我们在遍历过程中一遇到右括号，就应该和left数组的尾部元素匹配，然后将left数组的尾部元素删除。(其实模拟了一个栈过程)，此时我们知道尾部元素在S中的下标和右括号在S中的下标，不难得出这之间有多少个左括号。

  简单的参考代码如下：

 #include<cstdio>

#include<cstring>

using namespace std;

const int maxn = 50;

 

int main()

{

    char s[maxn];

    int p[maxn];

    int w[maxn];

    int t,index;

    scanf("%d",&t);

    while(t--)

    {

        int n;

        scanf("%d",&n);

        for(int i = 1;i <= n;i++)

              scanf("%d",&p[i]);

        index = 0;

        p[0] = 0;

 

        for(int i = 1;i <= n;i++)

        {

 

             int temp = p[i] - p[i-1];

 

              while(temp--)

              {

                  s[index++] = '(';

              }

                 s[index++] = ')';

 

        }   //得到s序列

 

 

        n *= 2;//得到w序列

        index = 1;

        int left[maxn];

        bool left2[ maxn];

        memset(left2 ,false , sizeof(left2));

 

        for(int i = 0;i < n;i++)

        {

 

 

              if(s[i] == '(')

                    left[index++] = i;

              else

                 {

                   if(i == n-1) {printf("%d
" ,(i-left[--index]+1)/2);}

                   else      {printf("%d ",(i-left[--index]+1)/2);}

                 }

        }

 

 

 

    }

}