poj 1068 Parencodings

最新推荐文章于 2022-10-09 16:11:21 发布

_txg

最新推荐文章于 2022-10-09 16:11:21 发布

阅读量270

点赞数 1

CC 4.0 BY-SA版权

分类专栏： ACM # ACM-算法设计

本文链接：https://blog.youkuaiyun.com/txgANG/article/details/52040276

ACM-算法设计同时被 2 个专栏收录

101 篇文章

订阅专栏

ACM

22 篇文章

订阅专栏

本文介绍了一种算法，用于将括号字符串的P序列（每个右括号左侧的左括号数量）转换为W序列（每个右括号内包含的括号数）。通过实例演示了如何使用整数数组表示括号并模拟转换过程。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

题目链接：点击打开链接

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：p序列是指每一个右括号左边有几个左括号，w序列是指每一个右括号包含几个括号

基本思路：先根据w序列还原，用int数组表示左右括号，早模拟

<span style="font-size:18px;">#include <iostream>
#include<cstring>
#include<stack>
#include<cstdio>
using namespace std;

int main()
{
    int a[100];
    int b[100];
    stack<int >q;
    int n,t,k;
    scanf("%d",&t);
    while(t--)//左1右2
    {

        scanf("%d",&n);
        int top=0;
        int x,y=0;
        for(int i=0; i<n; i++)
        {
            scanf("%d",&x);
            k=x-y;
            y=x;
            while(k--)
            {
                a[top++]=1;
            }
            a[top++]=2;
        }
        int p;
        int sum=0;
        int v=0;
        for(int i=0; i<top; i++)
        {
            sum=0;
            if(a[i]==2)
            {
                q.push(a[i]);
                p=i-1;
                while(!q.empty()&&p>=0)
                {
                    if(q.top()!=a[p])
                    {
                        sum++;
                        q.pop();
                    }
                    else q.push(a[p]);
                    p--;
                }
                b[v++]=sum;
            }
        }
        for(int i=0; i<v; i++)
        {
            if(i==0)printf("%d",b[i]);
            else printf(" %d",b[i]);
        }
        printf("\n");
    }
    return 0;
}
</span>