I - Parencodings

I - Parencodings

题目描述：

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.

题目大意：

这道题讲的是给出一串括号，然后有2种编码方式，分别为W与P，假设括号为(((()()())))，如果按照P编码来输出，就从第一个右括号他的左边有多少个括号就输出，接着按照第二个右括号的左边有多少个括号输出，则有4 5 6666。如果按照W编码来输出，就从 第一个右括号与他最近的左括号结合，输出2者的距离，但不能重复结合，一旦这个左括号与右括号结合，后面的右括号就不能与其在结合，就有 1 1 1 4 5 6.。

题目要求给出P编码的输出，然后叫你转换为W编码输出。

思路分析：

可以先将P编码打出括号字符串，然后在转换为W，则就相当于先进栈后出栈，并给特定的右括号做标记，就可以了。

代码块：

#include<stdio.h>
int  lemon()
{
	char str[2000];//记录字符串 
	int L[2000]={0};//判断左括号是否被标记了 
	int L1[2000];//记录W编码方式的结果。 
	int b,c,num=0,d,e,h=0;
	scanf("%d",&b);
	for(c=0;c<b;c++)//将P编码方式转为字符串 
	{
		scanf("%d",&d);
		for(e=0;e<d-num;e++)
		{
			str[h++]='('; 
		}
		num=d;
		str[h++]=')';
	}
	num=0;
	for(c=0;c<h;c++)//由字符串转化为W编码方式 
	{
		int num1=1;//左右括号之间的距离 
		if(str[c]==')')
		{
			for(e=c-1;e>=0;e--)
			{
				if(str[e]=='(' && L[e]==0)//当左括号被人结合后，L[e]就会被标记相当于
				{						//相当于给左括号下标做标记。 
					L[e]=1;
					L1[num++]=num1;//如果结合成功就将其2括号之间的距离记录下来。 
					break;	
				}
				else if(L[e]==1 && str[e]=='(')
				{
					num1++;//计算右括号与左括号之间的距离 
				}
			}
		}
	}
	for(c=0;c<num;c++)//输出W编码方式 
	{
		if(c!=num-1)
		printf("%d ",L1[c]);
		else
		printf("%d\n",L1[c]);
	}
}
int main()
{
	int b,c,d;
	scanf("%d",&b);
	for(c=0;c<b;c++)
	{
		lemon();
	}
}