Parencodings 分类： POJ 2015-06-28 22:00 7人阅读 评论(0) 收藏

Parencodings

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 22757		Accepted: 13337

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
#include <iostream>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <string>
#include <stack>
#include <algorithm>
using namespace std;

const int Max=1100000;

int Arr[110];
bool vis[1100];
int main()
{
    int T;
    int n;
    while(~scanf("%d",&T))
    {
        while(T--)
        {
            scanf("%d",&n);
            int Max=0;
            int sum;
            for(int i=0;i<n;i++)
            {
                scanf("%d",&Arr[i]);
                if(Arr[i]>Max)
                {
                    Max=Arr[i];
                }
            }
            Max+=n;
            sum=Max;
            memset(vis,false,sizeof(vis));
            vis[Max]=true;
            for(int i=n-2;i>=0;i--)
            {

                Max-=(Arr[i+1]-Arr[i]+1);
                vis[Max]=true;
            }
            int top=0;
            for(int i=1;i<=sum;i++)
            {
                if(vis[i])
                {
                    int ans=0;
                    int ant=0;
                    for(int j=i;j>=1;j--)
                    {
                        if(vis[j])
                        {
                            ans++;
                        }
                        else
                        {
                            ans--;
                            ant++;
                        }
                        if(!ans)
                        {
                            break;
                        }
                    }
                    Arr[top++]=ant;
                }
            }
            for(int i=0;i<n;i++)
            {
                if(i)
                    cout<<" ";
                cout<<Arr[i];
            }
            cout<<endl;
        }
    }
    return 0;
}

版权声明：本文为博主原创文章，未经博主允许不得转载。