HDU 5510 Bazinga（KMP||strstr神器+些许优化）

Ladies and gentlemen, please sit up straight. 
Don't tilt your head. I'm serious. 

For nn given strings S1,S2,⋯,SnS1,S2,⋯,Sn, labelled from 11 to nn, you should find the largest i (1≤i≤n)i (1≤i≤n) such that there exists an integer j (1≤j<i)j (1≤j<i) and SjSj is not a substring of SiSi. 

A substring of a string SiSi is another string that occurs in SiSi. 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)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)n (1≤n≤500) and in the following nn lines list are the strings S1,S2,⋯,SnS1,S2,⋯,Sn. 
All strings are given in lower-case letters and strings are no longer than 20002000letters.

Output

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

Sample Input

45ababczabcabcdzabcd4youlovinyouaboutlovinyouallaboutlovinyou5dedefabcdabcdeabcdef3abaccc

Sample Output

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

题解：

题意：

让你从一堆有编号的串中找出最后面的一个i，使得在[1，i-1]中可以找到一个串s[j]不是s[i]的子串

思路：

直接搞看上去都是要炸的，所以要做一些优化，比如如果s[j-1]是s[j]的子串，那么如果s[j-1]不是s[i]的子串，那么s[j]也不是，反之如果s[j]是s[i]的子串，那么s[j-1]也一定是s[i]的子串，这样想预处理扫一遍数组，将前后有这种包含关系的标记一下就好了，然后从后面往前暴力扫

代码：

#include<iostream>#include<cstring>#include<stdio.h>#include<math.h>#include<string>#include<stdio.h>#include<queue>#include<stack>#include<map>#include<vector>#include<deque>#include<algorithm>#define ll long long#define INF 1008611111#define M (t[k].l+t[k].r)/2#define lson k*2#define rson k*2+1using namespace std;char s[505][2005];int vis[505];int main(){    int i,j,k,test,q,n,d;    scanf("%d",&test);    for(q=1;q<=test;q++)    {        scanf("%d",&n);        for(i=1;i<=n;i++)        {            scanf("%s",s[i]);            vis[i]=0;        }        printf("Case #%d: ",q);        d=-1;        for(i=2;i<=n;i++)        {            if(strstr(s[i],s[i-1])!=NULL)                vis[i-1]=1;        }        for(i=n;i>=1;i--)        {            for(j=1;j<i;j++)            {                if(vis[j])                    continue;                if(strstr(s[i],s[j])==NULL)                {                    d=i;                    goto loop;                }            }        }        loop:;        printf("%d\n",d);    }    return 0;}