poj 1068 Parencodings (模拟)

Parencodings

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 17169		Accepted: 10296

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

 

题意：

对于给出的原括号串，存在两种数字密码串：

1.p序列：当出现匹配括号对时，从该括号对的右括号开始往左数，直到最前面的左括号数，就是pi的值。

2.w序列：当出现匹配括号对时，包含在该括号对中的所有右括号数（包括该括号对）,就是wi的值。

 

题目的要求：对给出的p数字串，求出对应的s串。串长均<=20.

提示：在处理括号序列时可以使用一个小技巧，把括号序列转化为01序列，左0右1，处理时比较方便

 

代码：

#include<cstdio>
#include<iostream>
#include<cstring>
using namespace std;

int p[30];
int w[30];
int s[100];

int main()
{
    int t,n,i,j,k,q;
    int sum1,sum2,head,rear,t1,t2,headw,rearw;
    scanf("%d",&t);
    while(t--)
    {
        sum1=0,sum2=0,head=0,rear=0;
        scanf("%d",&n);
        for(i=0;i<n;i++)
        {
            scanf("%d",&p[i]);
            if(sum1<p[i])
            {
                for(j=0;j<p[i]-sum1;j++)
                    s[rear++]=0;
                sum1=p[i];
            }
            s[rear++]=1;
            sum2++;
        }
        rearw=headw=0;
        while(head!=rear)
        {
            t1=0,t2=0;
            if(s[head]==1)
            {
                t1++;
                if(s[head-1]==0)
                    w[rearw++]=1;
                else
                {
                    t1++;
                    q=head-1;
                    while(t1!=t2)
                    {
                        q--;
                        if(s[q]==0)
                            t2++;
                        else
                            t1++;
                    }
                    w[rearw++]=t1;
                }
                head++;
            }
            else
                head++;
        }
        for(k=0;k<rearw;k++)
            printf("%d%c",w[k],k==rearw-1?'\n':' ');
    }
    return 0;
}

 

11924371

fukan

1068

Accepted

144K

0MS

C++

1426B

2013-08-05 16:19:34