poj 1035

Description

Input

Output

Sample Input

i
is
has
have
be
my
more
contest
me
too
if
award
#
me
aware
m
contest
hav
oo
or
i
fi
mre
#

Sample Output

me is correct
aware: award
m: i my me
contest is correct
hav: has have
oo: too
or:
i is correct
fi: i
mre: more me

Source

Northeastern Europe 1998

直接暴力了

AC代码：

#include<iostream>
#include<map>
#include<string>
#include<cstring>
#include<algorithm>
#include<stdio.h>
using namespace std;
string str[10010];
int judge1(string s1,string s2){
    int c,i; c=0;
    for(int i=0;i<s1.length() && i+c<s1.length();){
        if(s1[i+c]==s2[i])
            i++;
        else
            c++;
    }
    return 2-c;
}
int judge2(string s1,string s2){
    int i,c;
    c=0;
    for(i=0;i<s1.length();i++)
        if(s1[i]!=s2[i])
            c++;
    return 2-c;
}
int main(){
    string s;
    int i,j; j=0;
    while(cin>>s && s[0]!='#'){
        str[j++]=s;
    }
    while(cin>>s && s[0]!='#'){
        for(i=0;i<j;i++){
            if(str[i]==s){
                cout<<s<<" is correct"<<endl;
                break;
            }
        }
        if(i>=j){
            cout<<s<<':';
            for(i=0;i<j;i++){
                if(str[i].length()==s.length()){
                    if(judge2(s,str[i])==1){
                        cout<<' '<<str[i];
                    }
                }
                else{
                    if(str[i].length()>s.length()){
                        if(judge1(str[i],s)==1){
                            cout<<' '<<str[i];
                        }
                    }
                    else{
                        if(judge1(s,str[i])==1){
                            cout<<' '<<str[i];
                        }
                    }
                }
            }
            cout<<endl;
        }
    }
    return 0;
}

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poj 1035

原文：http://blog.csdn.net/my_acm/article/details/38345667