HDU5510 Bazinga （strstr()）

Bazinga

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 4919    Accepted Submission(s): 1552

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

45ababczabcabcdzabcd4youlovinyouaboutlovinyouallaboutlovinyou5dedefabcdabcdeabcdef3abaccc

 

Sample Output

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

 

Source

2015ACM/ICPC亚洲区沈阳站-重现赛（感谢东北大学）

思路：用strstr匹配字符串，假设存在两个字符串i，j，（i < j），如果第i个字符串是第j个字符串的子串，则直接跳出循环即可，因为接下来完全可以用第j个字符串来代替第i个字符串，如果第i个字符串不是第j个字符串的子串，则标记第j个字符串，因为第j个字符串已经是我们要寻找的字符串，以后不用再匹配。

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;int t,n,vis[510],cas=0;char str[510][2010];int main(){    scanf("%d",&t);    while(t --){        scanf("%d",&n);        for(int i = 1; i <= n; i ++) scanf("%s",str[i]);        memset(vis,0,sizeof(vis));        int ans = -1;        for(int i = 1; i <= n; i ++){            for(int j = i+1; j <= n; j ++){//如果第j个是第i个字符的母串，则没有必要再接着匹配下去                if(!vis[j]){                    if(strstr(str[j],str[i])) break;                    else{//否则标记第j个字符串即可                        vis[j] = 1; ans = max(ans,j);                    }                }            }        }        printf("Case #%d: %d\n",++cas,ans);    }    return 0;}