zoukankan      html  css  js  c++  java
  • 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;
    }
    



  • 相关阅读:
    DFC-3C和DFC-3B的区别和注意事项
    Bug搬运工-CSCux99539:Intermittent error message "Power supply 2 failed or shutdown"
    EVE上传Dynamips、IOL和QEMU镜像
    EVE扩大虚拟内存
    EVE磁盘扩容
    VMware安装EVE
    介绍Mobility Group
    Bug搬运工-CSCvi02106 :Cisco 2800, 3800, 1560 APs: when connected to a Cisco Switch CDP-4-DUPLEX_MISMATCH log is seen
    jquery.autocomplete在火狐下的BUG解决
    nodeJS中exports和mopdule.exports的区别
  • 原文地址:https://www.cnblogs.com/wgwyanfs/p/7189827.html
Copyright © 2011-2022 走看看