Parenthese sequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)

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

Author

Xiaoxu Guo (ftiasch)

Source

2014 Multi-University Training Contest 5

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

const int maxn=1001000;

char str[maxn];
int num[maxn],prefix[maxn],suffix[maxn];
int pre0[maxn],suf0[maxn];

void init()
{
    memset(num,0,sizeof(num));
    memset(prefix,0,sizeof(prefix));
    memset(suffix,0,sizeof(suffix));
    memset(pre0,0,sizeof(pre0));
    memset(suf0,0,sizeof(suf0));
}

int main()
{
while(scanf("%s",str)!=EOF)
{
    int n=strlen(str);
    if(n%2==1)
    {
        puts("None"); continue;
    }
    init();
    for(int i=0;i<n;i++)
    {
        if(str[i]=='(') num[i+1]=1;
        else if(str[i]==')') num[i+1]=-1;
        else if(str[i]=='?') num[i+1]=0;
    }
    bool flag=true;
    for(int i=1;i<=n;i++)
    {
        if(num[i]) prefix[i]=prefix[i-1]+num[i];
        else prefix[i]=prefix[i-1]+1;
        if(prefix[i]<0)
        {
            flag=0; break;
        }
    }
    for(int i=n;i>=1;i--)
    {
        if(prefix[i]<=1)
            pre0[i]=pre0[i+1]+1;
        else pre0[i]=pre0[i+1];
    }
    if(flag==false)
    {
        puts("None"); continue;
    }
    for(int i=n;i>=1;i--)
    {
        if(num[i]) suffix[i]=suffix[i+1]-num[i];
        else suffix[i]=suffix[i+1]+1;
        if(suffix[i]<0)
        {
            flag=false;
            break;
        }
    }
    for(int i=1;i<=n;i++)
    {
        if(suffix[i]<=1)
            suf0[i]=suf0[i-1]+1;
        else suf0[i]=suf0[i-1];
    }
    if(flag==false)
    {
        puts("None"); continue;
    }
    int cnt=0;
    for(int i=2;i<n;i++)
    {
        if(num[i]==0)
        {
            if( (prefix[i]>=2&&pre0[i]==0) && (suffix[i]>=2&&suf0[i]==0) )
                cnt++;
        }
    }
    if(cnt)
        puts("Many");
    else
        puts("Unique");
}
	return 0;
}

HDOJ 4915 Parenthese sequence