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  • HDOJ 4915 Parenthese sequence



    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 s1s2…sn (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
     



    #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;
    }
    



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  • 原文地址:https://www.cnblogs.com/wgwyanfs/p/7189827.html
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