字符串匹配

      对于两个字符串之间的匹配：

二分匹配实在看不懂，我自己的办法是设两个数组，用于标记字符串中'a'-'z'的个数，然后比较两个数组时候一致，一致则匹配

TOJ 1856.   Anagrammatic Distance 

统计两个字符串中字符的差异数

#include<iostream>
using namespace std;
int main()
{
    int test,i,j=1,k,dis,p1[27],p2[27];
    char s1[50],s2[50];
    cin>>test;
    cin.getline(s1,50);
    while(test--)
    {            
        cin.getline(s1,50);
        cin.getline(s2,50);
        dis=0;
        memset(p1,0,sizeof(p1));
        memset(p2,0,sizeof(p2));
        for(i=0;i<strlen(s1);i++)
        {k=s1[i]-'a';p1[k]++;}        
        for(i=0;i<strlen(s2);i++)
        {k=s2[i]-'a';p2[k]++;}        
        for(i=0;i<26;i++)
            dis+=abs(p1[i]-p2[i]);
        cout<<"Case #"<<j++<<":  ";
        cout<<dis<<endl;
    }
    return 0;
}

TOJ 2468.   Counting Letters

 统计字符串中各字符的数量

#include<iostream>
using namespace std;
int main()
{
    char text[101],e;
    int test,i,k,len,count[26],max;
    cin>>test;
    while(test--)
    {
        memset(count,0,sizeof(count));
        cin>>text;
        len=strlen(text);
        for(i=0;i<len;i++)
        {
            k=text[i]-'a';
            count[k]++;
        }
        max=0;
        for(i=0;i<26;i++)
            if(count[i]>max) max=count[i];
        for(i=0;i<26;i++)
            if(count[i]==max)
            {    e='a'+i;cout<<e;}
            cout<<endl;
    }
    return 0;
}

TOJ 2943.   Love Letter

此题极度纠结，输入麻烦，匹配也麻烦，写了半天仍然WA

后来Roba大牛告诉我“这个程序很怪，我把它拿到TOJ上去测，结果都是对的。不知道为什么交上去就是WA... ”

囧

#include<iostream>
using namespace std;
int i,j,p,q,n,k,t,o;
char s[100];
char l[30];
char m[100][30];
int p1[26],p2[26];


void convert(char l[21],int i)
{    
    if(i==2)
    {
        int find=0;
        for(p=0;p<k&&find==0;p++)
        {            
            if(l[0]==m[p][0]&&l[1]==m[p][1])
            {printf("%s",m[p]);find=1;break;}
        }        
    }    
    else{    
        int find=0;    
        for(p=0;p<k&&find==0;p++)
        {
            o=0;
            if(strlen(m[p])==i)
            {                
                if(l[0]==m[p][0]&&l[i-1]==m[p][i-1])
                {
                    memset(p1,0,sizeof(p1));
                    memset(p2,0,sizeof(p2));
                    for(q=1;q<i-1;q++)
                    {t=m[p][q]-'a';p1[t]++;}                    
                    for(q=1;q<i-1;q++)
                    {t=l[q]-'a';p2[t]++;}                    
                    for(;o<26;o++)
                        if(p1[o]!=p2[o])
                            break;                    
                }
            }
            if(o==26)
                    {printf("%s",m[p]);find=1;}
        }
    }    
}

int main()
{        
    cin>>n;
    k=n;
    i=0;
    while(n--)
    {
        cin>>m[i];
        i++;
    }
    
    cin.getline(s,100);//纯粹为了输出格式    
    while(cin.getline(s,100))
    {
        i=0;j=0;
        while(j<strlen(s))
        {
            if(s[j]==' ')
            {j++;convert(l,i);cout<<' ';i=0;}
            else 
            {
                l[i++]=s[j++];    
            }
        }
        convert(l,i);        
        cout<<endl;        
    }
    return 0;
}

TOJ 2628.   Subsequence

简单的字符串匹配，子序列 

#include<iostream>
#include<string>
using namespace std;
int main()
{
    int t,i,j,d1,d2;
    char s1[101],s2[101];
    cin>>t;
    while(t-->0)
    {
        cin>>s1>>s2;
        i=0;j=0;d1=0;d2=0;
        for(i=0;i<strlen(s1);i++)
        {
            if(s1[i]==s2[j])
                j++;
            if(j==strlen(s2))
                break;
        }
        if(j==strlen(s2))
            d1=1;
        j=strlen(s2)-1;
        for(i=0;i<strlen(s1);i++)
        {
            if(s1[i]==s2[j])
                j--;
            if(j==-1)
                break;
        }
        if(j==-1)
            d2=1;
        if(d1==1||d2==1)
            cout<<"YES"<<endl;
        else
            cout<<"NO"<<endl;
    }
    return 0;
}

TOJ 1911.   Easier Done than Said?

简单地统计字符数，判断单词是否符合语法规则

#include<iostream>
using namespace std;
int main()
{
    char s[21];
    int ac,i,v,c;
    while(cin>>s&&strcmp(s,"end")!=0)
    {
        ac=0;v=0;c=0;
        for(i=0;i<strlen(s);i++)
            if(s[i]=='a'||s[i]=='e'||s[i]=='i'||s[i]=='o'||s[i]=='u')
            {
                ac=1;
                break;
            }
        if(ac==1)
        {
        for(i=0;i<strlen(s);i++)
        {
            if(i>0&&s[i]==s[i-1]&&s[i]!='e'&&s[i]!='o')
            {
                ac=0;
                break;
            }
            if(s[i]=='a'||s[i]=='e'||s[i]=='i'||s[i]=='o'||s[i]=='u')
            {    v++;c=0;}
            else
            {   c++;v=0;}
            if(v==3||c==3)
            {   ac=0;break;}
        }
        }
        if(ac==1)
            cout<<"<"<<s<<"> is acceptable."<<endl;
        else 
            cout<<"<"<<s<<"> is not acceptable."<<endl;
    }
    return 0;
}