HDU 5510 Bazinga 多种姿势

Bazinga

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 4910    Accepted Submission(s): 1543

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）

 

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这个题比较好玩啊，感觉题意和图片没啥关系，可能是用来活动气氛的把，题意就是让你找到第i个字符串，他位置前面的字符串有一个不是他子串的字符串。

思路：已第i个串为基地，去找，如果他是他后面某一个串的子串，那么就break掉就可以了，否则，他后面的这个串就有可能是最后解，用max取一下最优

my ugly code(KMP匹配)

#include <cstdio>#include <cmath>#include <cstring>#include <string>#include <algorithm>using namespace std;int t,n;char a[521][2017];int cas=1;int e[521];void getfail(char *p,int *f){    int m=strlen(p);    f[0]=f[1]=0;    for(int i=1;i<m;i++){        int j=f[i];        while( j && p[j]!=p[i]) j=f[j];        f[i+1]=(p[i]==p[j])?j+1:0;    }}int kmp(char *t,char *p,int *f){    int n=strlen(t),m=strlen(p);    getfail(p,f);    int j=0;    for(int i=0;i<n;i++){        while(j && p[j]!=t[i]) j=f[j];        if(p[j] == t[i] ) j++;        if(j==m){            return 1;        }    }    return -1;}int main(){    int f[2017];    scanf("%d",&t);    while(t--){        scanf("%d",&n);        int max_len=0;        for(int i=1;i<=n;i++){            scanf("%s",a[i]);        }        for(int i=0;i<=n;i++)            e[i]=1;        int ans=-1,flag;        for(int i=1;i<n;i++){            for(int j=i+1;j<=n;j++){                if(e[j]){                    int tmp=kmp(a[j],a[i],f);                    if( tmp == 1){                        break ;                    }                    else{                        e[j]=0;                        ans=max(ans,j);                    }                }            }        }        printf("Case #%d: %d\n",cas++,ans);    }    return 0;}

my ugly code(strstr()匹配)

#include <cstdio>#include <cmath>#include <cstring>#include <string>#include <algorithm>using namespace std;int t,n;char a[521][2017];int cas=1;int e[521];void getfail(char *p,int *f){    int m=strlen(p);    f[0]=f[1]=0;    for(int i=1;i<m;i++){        int j=f[i];        while( j && p[j]!=p[i]) j=f[j];        f[i+1]=(p[i]==p[j])?j+1:0;    }}int kmp(char *t,char *p,int *f){    int n=strlen(t),m=strlen(p);    getfail(p,f);    int j=0;    for(int i=0;i<n;i++){        while(j && p[j]!=t[i]) j=f[j];        if(p[j] == t[i] ) j++;        if(j==m){            return 1;        }    }    return -1;}int main(){    int f[2017];    scanf("%d",&t);    while(t--){        scanf("%d",&n);        int max_len=0;        for(int i=1;i<=n;i++){            scanf("%s",a[i]);        }        for(int i=0;i<=n;i++)            e[i]=1;        int ans=-1,flag;        for(int i=1;i<n;i++){            for(int j=i+1;j<=n;j++){                if(e[j]){                    if( strstr(a[j],a[i]) ){                        break ;                    }                    else{                        e[j]=0;                        ans=max(ans,j);                    }                }            }        }        printf("Case #%d: %d\n",cas++,ans);    }    return 0;}

这里有个东西不明白，为什么strstr（）函数比kmp还快，如果有大佬看到这篇文章，或者你旁边有位大佬，求解答，感激不尽