Codeforces Round #282 (Div. 2) Treasure

原题链接：

http://codeforces.com/contest/495/problem/C

C. Treasure

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Malek has recently found a treasure map. While he was looking for a treasure he found a locked door. There was a string s written on the door consisting of characters '(', ')' and '#'. Below there was a manual on how to open the door. After spending a long time Malek managed to decode the manual and found out that the goal is to replace each '#' with one or more ')' characters so that the final string becomes beautiful.

Below there was also written that a string is called beautiful if for each i (1 ≤ i ≤ |s|) there are no more ')' characters than '(' characters among the first i characters of s and also the total number of '(' characters is equal to the total number of ')' characters.

Help Malek open the door by telling him for each '#' character how many ')' characters he must replace it with.

Input

The first line of the input contains a string s (1 ≤ |s| ≤ 105). Each character of this string is one of the characters '(', ')' or '#'. It is guaranteed that s contains at least one '#' character.

Output

If there is no way of replacing '#' characters which leads to a beautiful string print  - 1. Otherwise for each character '#' print a separate line containing a positive integer, the number of ')' characters this character must be replaced with.

If there are several possible answers, you may output any of them.

Sample test(s)

input

(((#)((#)

output

1
2

input

()((#((#(#()

output

2
2
1

input

#

output

-1

input

(#)

output

-1

Note

|s| denotes the length of the string s.

题意：一串字符串只含有'(',')','#',三种字符，你需要把其中的每个‘#’变为一个或者多个')'，使字符串满足两个条件，①整个字符串‘（’与‘）’数量相等，②任意位置i，在它和它之前的'）'数不能超过‘（’数。如果无解输出-1，否则每行输出从前往后每个‘#’所替换的‘）’量。

思路：由于并不要求任意位置都（）配对，所以可以把最后 一个‘#‘之前的’#‘都全部设置为1，只需要考虑最后一个’#‘是否满足条件即可。

方法：①首先保证每个‘#’至少替换一个‘）’，方法是，读入按单个字符读入，读入‘#’时，存储字符串增加’）‘’#’两个字符，并记录最后一个’#‘的位置last。②从开头往last扫，计算截止i一共多出多少个’（‘，记录为nl，若任意位置nl<0了，说明’）‘多于‘（’，而由于我们之前每个‘#’只置换一个‘）’，‘）’是最小值，则说明条件已经不满足。③从尾部往last扫，判断last之后是否出现不合理，这时候，同样记录‘（’的数量nr，如果任意位置nr>0，说明从i~n（总长度）中‘（’多于‘）’，相对应从1~i-1，就一定有‘（’多于‘）’，不满足条件。④从last扫往尾部，判断条件如①，但是此时正式的‘（’多出数量为nl+nr；

这道题应该有很简单的方法，但是太挫只会这个方法，还好数据量小。。。

#include "stdio.h"
#include "string.h"

int main()
{
	int n=0,nl=0,nr=0,ans[100010],nj=0,flag=1,last;
	char a[200010],in;
	memset(ans,0,sizeof(ans));
    while(1)
	{
		in=getchar();
		if(in=='\n')
		{
			break;
		}
		if(in=='#')
		{
			a[++n]=')';
			a[++n]='#';
			last=n;
		}
		else
			a[++n]=in;
	}
	for(int i=1;i<last;i++)
	{
		if(a[i]=='(')
		{
			nl++;
		}
		else if(a[i]==')')
		{
			nl--;
		}
		if(nl<0)
		{
			flag=0;
			break;
		}
	}
	if(flag)
	{
		for(int i=n;i>=last+1;i--)
		{
			if(a[i]=='(')
				nr++;
			else if(a[i]==')')
				nr--;
			if(nr>0)
			{
				flag=0;
				break;
			}
		}
	}
	nr=0;
	if(flag)
	{
		for(int i=last+1;i<=n;i++)
		{
			if(a[i]=='(')
				nr++;
			else if(a[i]==')')
				nr--;
			if(nl+nr<0)
			{
				flag=0;
				break;
			}
		}
	}
	if(flag)
	{
		for(int i=0;i<last;i++)
		{
			if(a[i]=='#')
				printf("1\n");
		}
		printf("%d\n",nl+nr+1);
	}
	else
	{
		printf("-1\n");
	}
	return 0;
}