POJ 1068-Parencodings（模拟-包含括号个数）

不念不忘少年蓝@ 2022-07-11 07:27 162阅读 0赞

Parencodings

<table> 
 <tbody> 
  <tr> 
   <td><strong>Time Limit:</strong>&nbsp;1000MS</td> 
   <td>&nbsp;</td> 
   <td><strong>Memory Limit:</strong>&nbsp;10000K</td> 
  </tr> 
  <tr> 
   <td><strong>Total Submissions:</strong>&nbsp;25510</td> 
   <td>&nbsp;</td> 
   <td><strong>Accepted:</strong>&nbsp;15016</td> 
  </tr> 
 </tbody> 
</table>

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][]

## 题目意思： ##

S		(((()()())))  括号表示
    
    P-sequence	4 5 6 6 6 6   每个右括号前面的左括号数目
    
    W-sequence	1 1 1 4 5 6   每对括号所包含的括号对数（包括其自身）

给出P序列，求其W序列。

## 解题思路： ##

必须先提醒一句，这题的测试数据中，n肯定大于20（虽然给的范围到20），一开始我开到21，直接RE…Orz

首先根据P序列还原出S序列：P\[i\]-P\[i-1\]是当前应在一个右括号前面插入的左括号数；

然后从前往后扫出未匹配的第一个右括号，再从当前位置往前扫出未匹配的第一个左括号；

最后在当前匹配的左右括号之间扫描已经匹配的括号数目，依次++即可。

#include<iostream>
    #include<cstdio>
    #include<cstdlib>
    #include<cstring>
    #include<algorithm>
    #include<cmath>
    using namespace std;
    struct N
    {
        char c;//存储'('或')'
        bool vis;//判断是否被访问
        int num;//c==')'时记录其包含的括号数
    } node[101];
    
    int a[101];//坑爹测试数据，开21果断RE…
    int main()
    {
        ios::sync_with_stdio(false);
        cin.tie(0);
        int t;
        cin>>t;
        while(t--)
        {
            int n;
            cin>>n;
            memset(a,0,sizeof(a));
            cin>>a[0];
            int cnt=0;
            for(int i=0; i<a[0]; ++i)//根据输入还原S（即括号表示）并初始化
            {
                node[cnt].c='(';
                node[cnt].vis=false;
                node[cnt++].num=0;
            }
            node[cnt].c=')';
            node[cnt].vis=false;
            node[cnt++].num=0;
            for(int i=1; i<n; ++i)
            {
                cin>>a[i];
                for(int j=0; j<a[i]-a[i-1]; ++j)
                {
                    node[cnt].c='(';
                    node[cnt].vis=false;
                    node[cnt++].num=0;
                }
                node[cnt].c=')';
                node[cnt].vis=false;
                node[cnt++].num=0;
            }
            /*for(int i=0; i<cnt; ++i)//输出S
                cout<<node[i].c<<" ";
            cout<<endl;*/
            bool flag=true;//判断当前是否仍有未匹配的括号
            while(flag)
            {
                flag=false;
                for(int i=1; i<cnt; ++i)//右括号‘)’
                    if(node[i].c==')'&&node[i].vis==false)
                        for(int j=i-1; j>=0; --j)//左括号‘(’
                            if(node[j].c=='('&&node[j].vis==false)
                            {
                                node[j].vis=node[i].vis=true;
                                node[i].num=1;
                                flag=true;
                                int k;
                                for(k=j; k<i; ++k)
                                    if(node[k].c==')'&&node[k].vis==true)
                                        ++node[i].num;//计算包含的括号数
                                break;
                            }
            }
            for(int i=0; i<cnt; ++i)
                if(node[i].c==')')
                    cout<<node[i].num<<" ";
            cout<<endl;
        }
        return 0;
    }
    /*
    2
    6
    4 5 6 6 6 6
    9
    4 6 6 6 6 8 9 9 9
    */

[Tehran 2001]: http://poj.org/searchproblem?field=source&key=Tehran+2001