hdoj 5510 Bazinga 【不要想太多。。。 strstr】

Bazinga Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 72    Accepted Submission(s): 28 Problem Description Ladies and gentlemen, please sit up straight. Don't tilt your head. I'm serious. 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：给你n个串，让你找到最大的i(字符串编号 从1开始)使得至少存在一个j(1<=j<i)满足串j不是串i的子串。

不要想太多，弱渣ac自动机TLE到死  o(╯□╰)o

AC代码：

#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <algorithm>
#include <map>
#include <queue>
#include <stack>
#include <string>
#include <vector>
#define lson o<<1|1, l, mid
#define rson o<<1, mid+1, r
#define ll o<<1
#define rr o<<1|1
#define INF 0x3f3f3f3f
#define eps 1e-8
#define debug printf("1\n")
#define MAXN 100010
#define MAXM 20000000
#define LL long long
#define CLR(a, b) memset(a, (b), sizeof(a))
#define W(a) while(a--)
#define Ri(a) scanf("%d", &a)
#define Pi(a) printf("%d\n", (a))
#define Rl(a) scanf("%lld", &a)
#define Pl(a) printf("%lld\n", (a))
#define Rs(a) scanf("%s", a)
#define Ps(a) printf("%s\n", (a))
#define MOD 1000000007
#define LL long long
using namespace std;
char str[510][2010];
int main()
{
    int t, kcase = 1;
    Ri(t);
    W(t)
    {
        int n; Ri(n);
        for(int i = 0; i < n; i++)
            Rs(str[i]);
        int ans = -2;
        for(int i = n-1; i > 0; i--)
        {
            if(!strstr(str[i], str[i-1]))
            {
                ans = max(ans, i);
                for(int j = i+1; j < n; j++)
                    if(!strstr(str[j], str[i-1]))
                        ans = max(ans, j);
            }
        }
        printf("Case #%d: %d\n", kcase++, ans+1);
    }
    return 0;
}