HDU4915--Parenthese sequence

Problem Description

bobo found an ancient string. The string contains only three charaters -- "(", ")" and "?".

bobo would like to replace each "?" with "(" or ")" so that the string is valid (defined as follows). Check if the way of replacement can be uniquely determined.

Note:

An empty string is valid.
If S is valid, (S) is valid.
If U,V are valid, UV is valid.

Input

The input consists of several tests. For each tests:

A string s₁s₂…s_n (1≤n≤10⁶).

Output

For each tests:

If there is unique valid string, print "Unique". If there are no valid strings at all, print "None". Otherwise, print "Many".

Sample Input

?? ???? (??

Sample Output

Unique
Many
None

思路：亂搞的、、、

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
#define maxn 1000800
char str[maxn],str2[maxn],S[maxn];
inline int min(int a,int b)
{
	return a>b?b:a;
}
int fuck(char * s)
{
	int left = 0,right = 0,res = 0;
	int len = strlen(s);
	if(len & 1)	return 0;
	bool flag = true;
	int last = 0;
	if(s[0] == ')')	return 0;
	else left++;
	for(int i = 1;i < len-1;i++)
	{
		if(s[i] == '(')	left++;
		else if(s[i] == ')')	right++;
		else res++;
		if(left+res < right)	return 0;
		if(left + res == right || left + res == right+1)	
		{
			last = i;//last之前全部確定
			left += res;
			res = 0;
		}
	}
	if(s[len-1] == '(')
		return 0;
	else 
	{
		right++;
		if(left + res == right)	last = len-1;
		if(right + res == left)	last = len-1;
	}
	if(left + res < right || right + res < left)	return 0;
	else if(last == len-1)	return 1;
	else return 2;
}
int main()
{
	//freopen("in.txt","r",stdin);
	while(scanf("%s",str)!=EOF)
	{
		int ans = 3;
		ans = min(ans,fuck(str));
		int len = strlen(str);
		for(int i = len-1;i >= 0;i--)//反過來也要跑一次。
		{
			char a;
			if(str[i] == '(')	a = ')';
			else if(str[i] == ')')	a = '(';
			else a = '?';
			str2[len-1-i] = a; 	
		}
		str2[len] = '\0';
		int hehe = fuck(str2);
		ans = min(ans,hehe);
		if(ans == 0)	cout << "None" << endl;
		else if(ans == 1)	cout << "Unique" << endl;
		else cout << "Many" << endl;
	}
}