暴力_______Rikka with Parenthesis II( hdu 5831 2016多校第八场)

Problem Description

As we know, Rikka is poor at math. Yuta is worrying about this situation, so he gives Rikka some math tasks to practice. There is one of them:

Correct parentheses sequences can be defined recursively as follows:

1.The empty string "" is a correct sequence.

2.If "X" and "Y" are correct sequences, then "XY" (the concatenation of X and Y) is a correct sequence.

3.If "X" is a correct sequence, then "(X)" is a correct sequence.

Each correct parentheses sequence can be derived using the above rules.

Examples of correct parentheses sequences include "", "()", "()()()", "(()())", and "(((())))".

Now Yuta has a parentheses sequence S,
and he wants Rikka to choose two different position i,j and
swap Si,Sj. 

Rikka likes correct parentheses sequence. So she wants to know if she can change S to a correct parentheses sequence after this operation.

It is too difficult for Rikka. Can you help her?

 

Input

The first line contains a number t(1<=t<=1000), the number of the testcases. And there are no more then 10 testcases with n>100

For each testcase, the first line contains an integers n(1<=n<=100000), the length of S. And the second line contains a string of length S which only contains ‘(’ and ‘)’.

 

Output

For each testcase, print "Yes" or "No" in a line.

 

Sample Input

3
4
())(
4
()()
6
)))(((

 

Sample Output

Yes
Yes
No

Hint
For the second sample input, Rikka can choose (1,3) or (2,4) to swap. But do nothing is not allowed.

 

题意：

给你一个只含有'('和')'的字符串问是否能让这个字符串交换两个字符后括号匹配？

分析：

思路简单，从左到右第一个富裕的 )  和从右到左第一个富裕的 ( 交换。然后判断是否匹配就可以了。注意这里有个问题就是假如字符串原本括号就是匹配的。如果要在交换一次。这里有个规律吧。如果字符串长度＞2那么一定存在一种交换方法能够使得交换后依然可以匹配。

这个题目细节很多。很容易wa.

代码：

#include <stdio.h>
#include <string.h>
int pre[100010],later[100010];
char str[100010];
int gethash(char c)
{
if(c == '(') return 1;
return -1;
}
int main()
{
int t,n,l,r;
scanf("%d",&t);
while(t--)
{
scanf("%d",&n);
scanf("%s",str+1);
n = strlen(str+1);
int flag = 0;
l = -1,r = n;
for(int i = 1 ; i <= n ; i ++)
flag += gethash(str[i]);
if(flag != 0)
{
printf("No\n");
continue;
}
for(int i = 1 ; i <= n ; i ++)
{
pre[i] = pre[i-1] + gethash(str[i]);
if(pre[i] < 0)
{
l = i;
break;
}
}
if(l == -1 )
{
if(n > 2) //排除掉 () 为 No 这组数据
printf("Yes\n");
else
printf("No\n");
continue;
}
later[n+1] = 0;
for(int i = n ; i >= 1 ; i --)
{
later[i] = later[i+1] + gethash(str[i]);
if(later[i] > 0)
{
r = i;
break;
}
}
if( r <= l )
{
printf("No\n");
continue;
}
char c = str[l];
str[l] = str[r];
str[r] = c;
for(int i = 1 ; i <= n ; i ++)
{
pre[i] = pre[i-1] + gethash(str[i]);
if(pre[i] < 0)
{
printf("No\n");
break;
}
if(i == n) printf("Yes\n");
}
}
return 0;
}


Problem Description

As we know, Rikka is poor at math. Yuta is worrying about this situation, so he gives Rikka some math tasks to practice. There is one of them:

Correct parentheses sequences can be defined recursively as follows:

1.The empty string "" is a correct sequence.

2.If "X" and "Y" are correct sequences, then "XY" (the concatenation of X and Y) is a correct sequence.

3.If "X" is a correct sequence, then "(X)" is a correct sequence.

Each correct parentheses sequence can be derived using the above rules.

Examples of correct parentheses sequences include "", "()", "()()()", "(()())", and "(((())))".

Now Yuta has a parentheses sequence S,
and he wants Rikka to choose two different position i,j and
swap Si,Sj. 

Rikka likes correct parentheses sequence. So she wants to know if she can change S to a correct parentheses sequence after this operation.

It is too difficult for Rikka. Can you help her?

 

Input

The first line contains a number t(1<=t<=1000), the number of the testcases. And there are no more then 10 testcases with n>100

For each testcase, the first line contains an integers n(1<=n<=100000), the length of S. And the second line contains a string of length S which only contains ‘(’ and ‘)’.

 

Output

For each testcase, print "Yes" or "No" in a line.

 

Sample Input

3
4
())(
4
()()
6
)))(((

 

Sample Output

Yes
Yes
No

Hint
For the second sample input, Rikka can choose (1,3) or (2,4) to swap. But do nothing is not allowed.

 

Author

学军中学