hdoj5510Bazinga【strstr+并查集】

Bazinga

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1438    Accepted Submission(s): 444

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）

 

刚做这道题时看别人给的一个思路是strstr以为很简单然后就用strstr暴力果断超时然后想到一个如果a是b的子串b是c的子串那么只需判断c和b即可然后想依靠这种思路优化。但不知道怎么实现。刚开始想的用vector存然后写出来超时。然后又用数组还是超时。最后猛然想到用并查集试试本来不报什么希望结果交了一发AC。

#include<cstdio>#include<cstdlib>#include<cstring>#include<algorithm>#include<cmath>#include<list>#include<queue>using namespace std;int father[510];char str[510][2010];int find(int x){return x==father[x]?x:father[x]=find(father[x]);}int main(){int n,i,j,k=1,t;scanf("%d",&t);while(t--){scanf("%d",&n);for(i=1;i<=n;++i){father[i]=i;}int ans=-1;for(i=1;i<=n;++i){scanf("%s",str[i]);for(j=i-1;j>=1;--j){int a=find(i),b=find(j);if(a==b)continue;else if(strstr(str[i],str[j])){father[i]=j;}else {ans=i;break;}}}printf("Case #%d: %d\n",k++,ans);}return 0;}