HDOJ 5510-Bazinga【模拟】

Bazinga

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 1468    Accepted Submission(s): 457

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）

解题思路：

就是找到一个最大的下标字符串，上面的字符串不是这一串的字串。

#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;char map[2000][2000];int main(){int T;int yy=0;scanf("%d",&T);while(T--){int n,i,j,k;scanf("%d",&n);for(i=1;i<=n;i++){scanf("%s",map[i]);}int ans=-1;for(i=n;i>1;i--){if(!strstr(map[i],map[i-1])){ans=max(ans,i);for(j=i-1,k=i+1;k<=n;k++){if(!strstr(map[k],map[j]))ans=max(ans,k);}}}printf("Case #%d: %d\n",++yy,ans);}return 0;}