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HDU 1361 & POJ 1068 Parencodings（模拟）

2024-07-25 08:30:08   217人阅读

题目链接：

HDU:http://acm.hdu.edu.cn/showproblem.php?pid=1361

POJ:http://poj.org/problem?id=1068


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：

有两个规则：

1、p序列表示一个括号字符串的每一个右括号的前面有多少个左括号；

2、w序列表示一个括号字符串的每一个右括号在和自己匹配的左括号之间（包括自己这一对）有多少了完整的括号；

现在给出p序列求w序列；

PS:

有序数据给的很小，所以直接模拟一下；


代码如下：

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int maxn = 47;
char s[maxn];
int w[maxn], p[maxn];
void solved_w(int n)
{
    int vis[maxn] = {0};
    int l = 0;
    for(int i = 0; i < 2*n; i++)
    {
        int k = 0;
        if(s[i] == ')')
        {
            for(int j = i-1; j >= 0; j--)
            {
                if(s[j] == '(')
                {
                    k++;
                }
                if(vis[j]==0 && s[j]=='(')
                {
                    vis[j] = 1;
                    w[l++] = k;
                    break;
                }
            }
        }
    }
    printf("%d",w[0]);
    for(int i = 1; i < n; i++)
    {
        printf(" %d",w[i]);
    }
    printf("\n");
}
int main()
{
    int t;
    int n;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        int f = 0, l = 0;
        for(int i = 0; i < n; i++)
        {
            scanf("%d",&p[i]);
            for(int j = 0; j < p[i]-f; j++)
            {
                s[l++] = '(';
            }
            s[l++] = ')';
            f = p[i];
        }
        //printf("%s\n",s);
        solved_w(n);
    }
    return 0;
}


HDU 1361 & POJ 1068 Parencodings（模拟）


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