POJ 1068 括号模拟

Parencodings

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 11191		Accepted: 6590

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-sequence4 5 6666

W-sequence1 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

先说说题目的意思吧，这道题是关于括号匹配的题目

首先输入一个t代表测试数据组数，紧接着输入n，代表右括号的数量

然后在输入n个数，代表每一个右括号的前面有多少个左括号

现在他的问题是，通过输入的数据判断，每一个右括号中间包括了多少对完整的括号

举个例子：(()())

第一个右括号中只括下了他自己一对括号，即1

第二个右括号和第一个情况一样，也是1

但是第三个右括号中间就扩了两个括号进去(()())

这道题，我最先开始想的是利用回溯算法来统计个数，可是怎么也没写对，杯具了很久之后想到直接统计右括号数量并且向前模拟的方法，然后很快就AC了。。后来仔细看了两份代码，发现，改的第一次写的代码中间已经用到了第二次写的一些思路，只不过大体的方向搞错了，所以一直有问题，不过从理论上讲，利用回溯法也应该可以AC，不过可能要麻烦许多。。。

AC代码：

Source Code

Problem: 1068		User: bingshen
Memory: 176K		Time: 0MS
Language: C++		Result: Accepted

• Source Code

  #include<stdio.h>
  #include<string.h>
  
  char s[100];
  int p[100];
  int right[100];
  int w[100];
  int ans[100];
  
  int main()
  {
  	int i,n,t,l,j,num;
  	scanf("%d",&t);
  	while(t--)
  	{
  		scanf("%d",&n);
  		memset(p,0,sizeof(p));
  		memset(w,0,sizeof(w));
  		memset(s,'(',sizeof(s));
  		memset(right,0,sizeof(right));
  		l=0;
  		num=0;
  		for(i=1;i<=n;i++)
  		{
  			scanf("%d",&p[i]);
  			if(i==1)
  				right[i]=p[i]+1;
  			else
  				right[i]=right[i-1]+(p[i]-p[i-1])+1;
  		}
  		for(i=1;i<=n;i++)
  			s[right[i]]=')';
  		int lc=0,rc=0;
  		for(i=1;i<=n;i++)
  		{
  			for(j=right[i];j>=1;j--)
  			{
  				if(s[j]=='(')
  					lc++;
  				if(s[j]==')')
  					rc++;
  				if(lc==rc)
  					break;
  			}
  			printf("%d ",rc);
  			lc=0;
  			rc=0;
  		}
  		printf("/n");
  	}
  	return 0;
  }

附一份用回溯写的Wrong Answer程序

Source Code



Problem: 1068



User: bingshen




Memory: N/A


Time: N/A



Language: C++


Result: Wrong Answer





Source Code 
#include<stdio.h>
#include<string.h>

char s[100];
int p[100];
int right[100];
int w[100];
int ans[100];

int dfs(int l,int r)
{
	int lc=0,rc=0,i,sum=0,start;
	if(l==r-1)
		return 1;
	else
	{
		start=l+1;
		for(i=l+1;i<=r-1;i++)
		{
			if(s[i]=='(')
				lc++;
			if(s[i]==')')
				rc++;
			if(lc==rc)
			{
				w[i]=dfs(start,i);
				sum=sum+w[i];
				lc=0;
				rc=0;
				start=i+1;
			}
		}
		return sum+1;
	}
}

int main()
{
	int i,n,t,l,j,num;
	scanf("%d",&t);
	while(t--)
	{
		scanf("%d",&n);
		memset(p,0,sizeof(p));
		memset(w,0,sizeof(w));
		memset(s,'(',sizeof(s));
		memset(right,0,sizeof(right));
		l=0;
		num=0;
		for(i=1;i<=n;i++)
		{
			scanf("%d",&p[i]);
			if(i==1)
				right[i]=p[i]+1;
			else
				right[i]=right[i-1]+(p[i]-p[i-1])+1;
		}
		for(i=1;i<=n;i++)
			s[right[i]]=')';
		w[right[n]]=dfs(1,right[n]);
		for(i=1;i<=right[n];i++)
		{
			if(w[i])
				ans[num++]=w[i];
		}
		if(num==0)
			printf("/n");
		for(i=0;i<num-1;i++)
			printf("%d ",ans[i]);
		printf("%d/n",ans[num-1]);
	}
	return 0;
}