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POJ1068_Parencodings

发表于: 2011-11-26   作者:Coco_young   来源:转载   浏览:
摘要: Parencodings Time Limit: 1000MS  Memory Limit: 10000K Total Submissions: 13268  Accepted: 7886 Description Let S = s1 s2...s2n be a well-formed string of parentheses. S can be encode
Parencodings
Time Limit: 1000MS  Memory Limit: 10000K
Total Submissions: 13268  Accepted: 7886

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

题目大意:一个完备的括号串(即所有括号都可以成功匹配)可以以2种编码方式编码
P型编码,在右括号出现的位置的值为,之前左括号出现的次数。
W型编码,在右括号出现的位置的值为,一个以该括号为终点的子完备括号串中右括号的个数。
输入一个括号串的P型编码,要求求出W型编码。

代码:
#include<iostream>
using namespace std;
int p[22];
int w[22];
int main()
{
    int t;
    cin>>t;
    int n;
    while(t--)
    {
        cin>>n;
        for(int i=1;i<=n;i++)
        cin>>p[i];
        w[0] = 0;
        p[0] = 0;
        for(int i=1;i<=n;i++)
        {
            int val = p[i]-p[i-1];
            if(val>0)
            {
                w[i] = 1;
            }
            else
            {
                int temp = 1;
                int cnt = 1;
                for(int j=i;j>=1;j--)
                {
                    temp--;
                    temp+=p[j]-p[j-1];
                    if(temp>0)
                    {
                        w[i] = cnt;
                        break;
                    }
                    else
                    cnt++;
                }
            }
        }
        for(int i=1;i<=n;i++)
        {
            if(i!=1)
            cout<<" "<<w[i];
            else
            cout<<w[i];
        }
        cout<<endl;
    }
    return 0;
}

POJ1068_Parencodings

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