HDU 5510 Bazinga(KMP)
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Bazinga
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 2986 Accepted Submission(s): 958
Problem Description
Ladies and gentlemen, please sit up straight.
Don't tilt your head. I'm serious.
Forn 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 stringSi is another string that occurs in Si . For example, ``ruiz" is a substring of ``ruizhang", and ``rzhang" is not a substring of ``ruizhang".
Don't tilt your head. I'm serious.
For
A substring of a string
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 integern (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 than2000 letters.
For each test case, the first line is the positive integer
All strings are given in lower-case letters and strings are no longer than
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
题意:有n个字符串S1,S2...Sn,要求找到最大的Si使得存在Sj不出现在Si中;
字符串匹配典型题,所以毫无疑问KMP(鉴于我还没有完全弄懂KMP是啥,所以直接在网上搜了个板子套上去了)
KMP的时间复杂度是O(n+m),结合题目的数据量,50个样例,500个字符串,每个字符串不超过2000,4000*500^2*50=5*10^8,直接KMP的话会TLE,因此在比较时对比较过程做一个优化;
对于一个字符串S1,如果它相匹配的字符串S2和另一个字符串S3做比较时,S3在遍历时已经毋须再与S1进行比较;因为如果S3与S2匹配,那么与S1肯定匹配;如果S3与S1不匹配,那刚好我们就可以认为3是我们暂时的答案;(因为这里的匹配是一种包含关系,S2包含S1,那么就相当于S3比较时也在和S2中包含的S1部分比较,以及S2的S2部分(即S2本身)比较)
其实我也不知道能优化到什么程度,反正不用每次都比较应该会省时~
然后一定要用scanf……不然会还是会TLE= =
可以在二重循环外输入,也可以在二重循环内中用i直接输入(这个还要快50ms);
#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int N1=505;const int N2=2005;int f[N2]; char str[N1][N2];//string str[N];int vis[N1];void getfill(string s) { memset(f,0,sizeof(f)); //根据其前一个字母得到 for(int i=1;i<s.size();i++) { int j=f[i]; while(j && s[i]!=s[j]) j=f[j]; f[i+1]=(s[i]==s[j])?j+1:0; } } int find(string a,string s) { int ans=0; getfill(s);int j=0; for(int i=0;i<a.size();i++) { while(j && a[i]!=s[j]) j=f[j]; if(a[i]==s[j]) j++; if(j==s.size()){ ans++; } } return ans; } int main(){int t,cas=1;scanf("%d",&t);while(t--){int n;scanf("%d",&n);int ans=-1;memset(vis,0,sizeof(vis));for(int i=1;i<=n;i++) scanf("%s",str[i]);for(int i=1;i<=n;i++){for(int j=1;j<=i-1;j++){if(vis[j]) continue;if(find(str[i],str[j])){vis[j]=1;}else ans=i;}}printf("Case #%d: %d\n",cas++,ans);}}/*45ababczabcabcdzabcd4youlovinyouaboutlovinyouallaboutlovinyou5dedefabcdabcdeabcdef3abaccc*/
接下来就是要弄懂KMP究竟是啥了~板子很好用啊~
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