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  • POJ-1068 Parencodings


    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

    水题。由于没注意空格错了一会。

    题意就是告诉第i个右括号前有几个左括号

    打印第i个括号,前有几个匹配的括号

    思路用的:STL队列,逆匹配

    #include <iostream>
    #include <cstdio>
    #include <cstring>
    #include <cstdlib>
    #include <algorithm>
    #include <queue>
    using namespace std;
    char s[1000];
    int main()
    {
        std::ios::sync_with_stdio(false);
        int t,n,i,j,a[30],b[30],l;
        cin>>t;
        while(t--)
        {
            cin>>n;
            l = 0;
            for(i = 0;i<n;i++)
                    {
                        cin>>a[i];
                        b[i] = a[i] + i;
                    }
            for( i = 0;i<a[0];i++)
                s[i] = '(';
            s[i] = ')';
            l = i;
            l++;
            for(i = 1;i<n;i++)
            {
                for(j = 0;j<a[i]-a[i-1];j++)
                    s[l++] = '(';
                s[l++] = ')';
            }
            s[l] = '';
            queue<char>q;
            int co = 0;
            for(i = 0;i<n;i++)
            {
                q.push(s[b[i]]);
                co = 0;
                for(j = b[i]-1;j>=0;j--)
                {
                    if(q.empty())
                        break;
                    if(s[j]==')')
                        q.push(s[j]);
                    else if(s[j]=='(' && q.front()==')')
                        {
                            q.pop();
                            co++;
                        }
                }
                if(i<n-1)
                printf("%d ",co);
                else
                    printf("%d
    ",co);
            }
        }
        return 0;
    }


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  • 原文地址:https://www.cnblogs.com/yutingliuyl/p/6820488.html
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