CodeForces 91A Newspaper Headline

題目描述

A newspaper is published in Walrusland. Its heading is s ₁ , it consists of lowercase Latin letters. Fangy the little walrus wants to buy several such newspapers, cut out their headings, glue them one to another in order to get one big string. After that walrus erase several letters from this string in order to get a new word s ₂ . It is considered that when Fangy erases some letter, there's no whitespace formed instead of the letter. That is, the string remains unbroken and it still only consists of lowercase Latin letters.

For example, the heading is "abc". If we take two such headings and glue them one to the other one, we get "abcabc". If we erase the letters on positions 1 and 5, we get a word "bcac".

Which least number of newspaper headings s ₁ will Fangy need to glue them, erase several letters and get word s ₂ ?

輸入描述:

The input data contain two lines. The first line contain the heading s₁, the second line contains the word s₂. The lines only consist of lowercase Latin letters (1 ≤ |s₁| ≤ 10⁴, 1 ≤ |s₂| ≤ 10⁶).

輸出描述:

If it is impossible to get the word s₂ in the above-described manner, print "-1" (without the quotes). Otherwise, print the least number of newspaper headings s₁, which Fangy will need to receive the word s₂.

示例1

輸入

複製

abc
xyz
abcd
dabc

輸出

複製

-1
2

題意：給出兩個字符串s1,s2，問須要幾個s1字符串才能獲得s2的序列（能夠是不連續的）

#include<iostream>
#include<algorithm>
#include<math.h>
#include<string.h>
#include<set>
#define ll long long
#define M 0x3f3f3f3f3f
using namespace std;
string s1,s2;
set<int>s[26];
int main()
{
    while(cin>>s1>>s2)
    {
        for(int i=0;i<26;i++)
            s[i].clear();
        int flag=0,cnt=1,last=-1;
        for(int i=0;s1[i];i++)
            s[s1[i]-'a'].insert(i);
        for(int i=0;s2[i];i++)
        {
            if(s[s2[i]-'a'].size()==0)
            {
                flag=1;
                break;
            }
            else
            {
                set<int>::iterator it=s[s2[i]-'a'].upper_bound(last);
                if(it==s[s2[i]-'a'].end())
                {
                    cnt++;
                    last=-1;
                }
                last=*s[s2[i]-'a'].upper_bound(last);
            }
        }
        if(flag==1)
            cout<<-1<<endl;
        else
            cout<<cnt<<endl;
        
    }
    return 0;
}