C++博客-Vontroy-随笔分类-字符串

字典树(Trie树)

Vontroy — Wed, 31 Aug 2016 01:35:00 GMT

如图，每条路径上保存一个字母，从根节点开始进行dfs，每遍历到一个标记节点（图中的红点），从根节点到当前节点路径上所有字母连起来即为一个单词

上图中存储了 abc, abcd, b, bcd, efg, hij.

对于Trie树主要有三种操作：

新建一个结点
插入一个单词
在trie树中查找单词

trie树中每次插入一个结点的时间复杂度是 O( strlen( str ) )

建立trie树的时间复杂度为 O( ∑strlen( str[i] ) )

对于每个单词，查询的时间复杂度为 O( strlen( str ) )

Templates:

Struct:

struct node
{
    int flag;     //The end of a word
    node* next[26];
};

新建结点：

node* newnode()
{
    node* p = new node;

    p->flag = 0;

    for( int i = 0; i < 26; i++ )
        p->next[i] = NULL;

    return p;
}

插入单词：

void trie_insert( node* root, char* s )
{
    node* p = root;
    int len = strlen( s );

    int tmp;
    for( int i = 0; i < len; i++ )
    {
        tmp = s[i] - 'a';
        if( p->next[tmp] == NULL )
            p->next[tmp] = newnode();
        p = p->next[tmp];
    }
    p->flag = 1;
}

查询：

int trie_search( node* root, char* s )
{
    node* p = root;
    int len = strlen( s );

    int tmp;
    for( int i = 0; i < len; i++ )
    {
        tmp = s[i] - 'a';
        if( p->next[tmp] == NULL )         //not matched
            return 0;
        p = p->next[tmp];
    }

    if( p->flag )     //match && it is a word which has been stored
return 1;
    return 0;
}

Vontroy 2016-08-31 09:35 发表评论

HDU 2734 Quicksum 简单字符串处理

Vontroy — Sun, 03 Oct 2010 02:03:00 GMT

Quicksum

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 492 Accepted Submission(s): 408

Problem Description

A checksum is an algorithm that scans a packet of data and returns a single number. The idea is that if the packet is changed, the checksum will also change, so checksums are often used for detecting transmission errors, validating document contents, and in many other situations where it is necessary to detect undesirable changes in data.

For this problem, you will implement a checksum algorithm called Quicksum. A Quicksum packet allows only uppercase letters and spaces. It always begins and ends with an uppercase letter. Otherwise, spaces and letters can occur in any combination, including consecutive spaces.

A Quicksum is the sum of the products of each character's position in the packet times the character's value. A space has a value of zero, while letters have a value equal to their position in the alphabet. So, A=1, B=2, etc., through Z=26. Here are example Quicksum calculations for the packets "ACM" and "MID CENTRAL":

ACM: 1*1 + 2*3 + 3*13 = 46MID CENTRAL: 1*13 + 2*9 + 3*4 + 4*0 + 5*3 + 6*5 + 7*14 + 8*20 + 9*18 + 10*1 + 11*12 = 650

Input

The input consists of one or more packets followed by a line containing only # that signals the end of the input. Each packet is on a line by itself, does not begin or end with a space, and contains from 1 to 255 characters.

Output

For each packet, output its Quicksum on a separate line in the output.

Sample Input

ACM
MID CENTRAL
REGIONAL PROGRAMMING CONTEST
ACN
A C M
ABC
BBC
#

Sample Output

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

int main()
{
    char str[260];
    int ans = 0;
    while( gets(str) && str[0] != '#' )
    {
        ans = 0;
        int len = strlen(str);
        for( int i = 0; i < len; i++ )
        {
            if( str[i] == ' ' )
                continue;
            else
                ans += ( str[i] - 64 ) * ( i + 1 );
        }
        std::cout << ans << std::endl;
    }
    return 0;
}

Vontroy 2010-10-03 10:03 发表评论

POJ 1007 DNA Sorting 字符串处理|稳定排序

Vontroy — Sat, 02 Oct 2010 13:23:00 GMT

/*****************
字符串处理
稳定排序
******************/

#include <iostream>
#include <algorithm>
#include <string>

using namespace std;

struct DNA
{
    int pos;
    int cnt;
    string str;
};

bool cmp(const DNA &a, const DNA &b)
{
    if (a.cnt != b.cnt)
    {
        return a.cnt < b.cnt;
    }
    else
    {
        return a.pos < b.pos;
    }
}

int main()
{
    int n, m, count;
    DNA ans[110];
    string str;
    cin >> n >> m;

    for (int i = 0; i < m; i++)
    {
        cin >> str;
        count = 0;

        for (int j = 0; j < n - 1; j++)
            for (int k = j + 1; k < n; k++)
                if (str[j] > str[k]) count++;

        ans[i].str = str;
        ans[i].cnt = count;
        ans[i].pos = i;
    }

    sort(ans, ans + m, cmp);

    for (int i = 0; i < m; i++)
        cout << ans[i].str << endl;

    return 0;
}

Vontroy 2010-10-02 21:23 发表评论

POJ 1002 487-3279 字符串处理

Vontroy — Sat, 02 Oct 2010 11:21:00 GMT

#include <iostream>
#include <string>
#include <algorithm>
#include <cstdio>

const int maxn = 20000;
const int MAXN = 100000+10;

using std :: string;
using std :: cout;
using std :: endl;

int main()
{
    string ans[MAXN];
    char tel[maxn];
    char store[maxn];
    int n;
    bool ok;
   while(~ scanf("%d",&n))
    {
        ok = false;
        int count = 0;
        while(n--)
        {
            scanf("%s",tel);
            int j;
            for(int i = 0,j = 0; tel[i] != '\0'; i++)
            {
                if(tel[i]-'0'>=0&&tel[i]-'0'<=9)
                      store[j++] = tel[i];
                else if(tel[i] != '-' && j != 3 )
                {
                    //ans[count]+='2';
                    if(tel[i]=='A'||tel[i]=='B'||tel[i]=='C')
                            store[j++] = '2';
                    else if(tel[i]=='D'||tel[i]=='E'||tel[i]=='F')
                            store[j++] = '3';
                    else if(tel[i]=='G'||tel[i]=='H'||tel[i]=='I')
                            store[j++] = '4';
                    else if(tel[i]=='J'||tel[i]=='K'||tel[i]=='L')
                            store[j++] = '5';
                    else if(tel[i]=='M'||tel[i]=='N'||tel[i]=='O')
                            store[j++] = '6';
                    else if(tel[i]=='P'||tel[i]=='R'||tel[i]=='S')
                            store[j++] = '7';
                    else if(tel[i]=='T'||tel[i]=='U'||tel[i]=='V')
                            store[j++] = '8';
                    else if(tel[i]=='W'||tel[i]=='X'||tel[i]=='Y')
                            store[j++] = '9';
                }
                if( j==3 )  store[j++]='-';
            }
            ans[count++] = store;
        }
        std :: sort(ans,ans+count);
        int ans_count;
        for(int i = 0; i < count; i++)
        {
             ans_count = 1;
             for(int j = i + 1; j < count; j++)
             {
                 if(ans[j]==ans[i]) ans_count++;
                 else break;
             }
             if(ans_count > 1)
             {
                 ok = true;
                 cout << ans[i] << " " << ans_count << endl;
                 i+=(ans_count-1);
             }
        }
        if(!ok)  cout << "No duplicates." << endl;

        //for(int i = 0; i         //std :: cout << ans[i] << std :: endl;
    }
    return 0;
}

Vontroy 2010-10-02 19:21 发表评论