HDOJ 4915 Parenthese sequence

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码代码的猿猿的AC之路

最新推荐文章于 2017-07-24 09:08:57 发布

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本文介绍了一个算法问题，即如何判断一个包含'(', ')' 和 '?' 的古代字符串是否能唯一地转换成有效的括号序列。有效括号序列需遵循特定规则：空串为有效，若S有效则(S)也有效，若U和V都有效则UV也有效。

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Parenthese sequence

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 716 Accepted Submission(s): 335

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 s ₁s ₂…s _n (1≤n≤10 ⁶).

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;
}