POJ 1068 Parencodings(模拟)

Parencodings

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 13696		Accepted: 8145

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

 解题报告：这道题的题意就是给我们P序列，让我们求S序列，P-sequence表示第i个‘)’之前有几个‘(’，而S-sequence是表示在第i个“）”和它相匹配的（号中有几对相匹配的“（）”的个数，包括它本身；现根据P序列求出括号的排列顺序，再求出S序列即可，主要就是找规律！

代码如下：

#include <iostream>
#include <cstdio>
#include <cstring>
const int N = 50;
int pse[N], wse[N], fuhao[N];
int visit[N];
int main()
{
    int t, n, i, j, k;
    scanf("%d", &t);
    while (t --)
    {
        memset(pse, 0, sizeof(pse));
        memset(wse, 0, sizeof(wse));
        memset(fuhao, 0, sizeof(fuhao));
        memset(visit, 0, sizeof(visit));
        scanf("%d", &n);
        for (i = 1; i <= n; ++i)
        {
            scanf("%d", &pse[i]);
        }
        int count;
        for (i = 1, count = 1; i <= n; ++i)
        {
            int num = pse[i] - pse[i -1];//两个相邻的有括号中间的左括号的个数
            while (num --)
            {
                fuhao[count ++] = 0;//符号为左括号时
            }
            fuhao[count ++] = 1;
        }
        k = 1;
        for (i = 1; i <= 2 * n; ++i)
        {
            if (fuhao[i])
            {
                count = 1;
                for (j = i - 1; j >= 1; -- j)
                {
                    if (visit[j]==0 && !fuhao[j])//找到与之相匹配的（的位置
                    {
                        visit[j] = 1;
                        break;
                    }
                    else if (visit[j] && !fuhao[j])//求中间的左括号（的个数
                    {
                        count ++;
                    }
                }
                wse[k ++] = count;
            }
        }
        for (i = 1; i < k - 1; ++i)
        {
            printf("%d ", wse[i]);
        }
        printf("%d\n", wse[i]);
    }
    return 0;
}