hdu 5510 strstr／find／KMP

HDU-5510-Bazinga

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Problem Description 
For n given strings S1,S2,⋯,Sn, labelled from 1 to n, you should find the largest i (1≤i≤n) such that there exists an integer j (1 ≤ j < i)and Sj is not a substring of Si.

A substring of a string Si is another string that occurs in Si. 
For example, “”ruiz” is a substring of “”ruizhang”“, and “”rzhang”“is not a substring of “”ruizhang”“.

Input 
The first line contains an integer t (1≤t≤50) which is the number of test cases. 
For each test case, the first line is the positive integer n (1≤n≤500) and in the following n lines list are the strings S1,S2,⋯,Sn. 
All strings are given in lower-case letters and strings are no longer than 2000 letters.

Output 
For each test case, output the largest label you get. If it does not exist, output −1.

Sample Input 
4 
5 
ab 
abc 
zabc 
abcd 
zabcd 
4 
you 
lovinyou 
aboutlovinyou 
allaboutlovinyou 
5 
de 
def 
abcd 
abcde 
abcdef 
3 
a 
ba 
ccc

Sample Output 
Case #1: 4 
Case #2: -1 
Case #3: 4 

Case #4: 3

题意：给N个串，求最大的i，使得存在j（<=0j<i），串j不是i的字串，有则输出i，否则暑促－1.

思路：先跑一遍循环，前后两个串进行比较，如果钱一个串是后一个串的字串，那钱一个串就没有用了。

eg：如果第i个串是第i+1个串的字串，如果第i＋1个串是i+1后变的串的字串，那么第i个串久没有用了，如果第i＋1个串不是后边串的字串，那么就找到了结果。

即定义指针i，j，如果s[i]是s［j］的字串，＝＝>i++

else  ans = j;

(因为如果s［i］是s［j］的字串，那么i＋＋，＝＝> 1~i-1全部的串一定是i～j中某些串的字串，所以i～l－1就没有必要和后边的串进行比较了)／／剪枝

如何判断s［j］是否是s［i］的字串？

1.KMP

2.s[i].find(s[j])!=string::npos

3.strstr(s[i],s[j])!=-1

其他解法：KMP＋二分

////  main.cpp//  160929////  Created by 刘哲 on 17/5/30.//  Copyright © 2016年 my_code. All rights reserved.//#include <bits/stdc++.h>#include <iostream>#include <algorithm>#include <cstdio>#include <cmath>#include <cstring>#include <string>#include <string.h>#include <map>#include <set>#include <queue>#include <deque>#include <list>#include <bitset>#include <stack>#include <stdlib.h>#define lowbit(x) (x&-x)typedef long long ll;typedef long long LL;using namespace std;int n;char s[505][2010];int flag[505];int main(){    int T;    cin>>T;    int n;    int cas = 1;    while(T--)    {        cin>>n;        for(int i=1;i<=n;i++)            cin>>s[i];        memset(flag,1,sizeof(flag));        int ans=-1;        for(int i=1;i<n;i++)        {            for(int j=i+1;j<=n;j++)            if(flag[j])            {                if(strstr(s[j],s[i]))break;                else                {                    flag[j]=0;                    ans=max(ans,j);                }            }        }        cout<<"Case #"<<cas++<<": "<<ans<<endl;    }    return 0;}

或者：

////  main.cpp//  160929////  Created by 刘哲 on 17/5/30.//  Copyright © 2016年 my_code. All rights reserved.//#include <bits/stdc++.h>#include <iostream>#include <algorithm>#include <cstdio>#include <cmath>#include <cstring>#include <string>#include <string.h>#include <map>#include <set>#include <queue>#include <deque>#include <list>#include <bitset>#include <stack>#include <stdlib.h>#define lowbit(x) (x&-x)typedef long long ll;typedef long long LL;using namespace std;int n;//char s[505][2010];string s[501];int flag[505];int main(){    int T;    cin>>T;    int n;    int cas = 1;    while(T--)    {        cin>>n;        for(int i=1;i<=n;i++)            cin>>s[i];        memset(flag,1,sizeof(flag));        int ans=-1;        for(int i=1;i<n;i++)        {            for(int j=i+1;j<=n;j++)            if(flag[j])            {                //if(strstr(s[j],s[i]))                if(s[j].find(s[i])!=string::npos)                    break;                else                {                    flag[j]=0;                    ans=max(ans,j);                }            }        }        cout<<"Case #"<<cas++<<": "<<ans<<endl;    }    return 0;}