hdu 5510 Bazinga 剪枝+find()/strstr()/KMP

Bazinga

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 3123    Accepted Submission(s): 1002

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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Statistic | Submit | Discuss | Note

三个方法不用剪枝似乎都会超时，

剪枝方法：加入a是b的子串，那么考虑c串是否能作为答案时，不用考虑a，只用考虑b。

因为如果b不是c的子串，c已经可以作为答案，自然不用考虑a。如果b是c的子串，那么a自然也是c的子串，并无任何帮助。

1.find()

#include<cstdio>#include<string>#include<cstring>#include<iostream>#include<cmath>#include<algorithm>#include<vector>using namespace std;#define all(x) (x).begin(), (x).end()#define for0(a, n) for (int (a) = 0; (a) < (n); (a)++)#define for1(a, n) for (int (a) = 1; (a) <= (n); (a)++)#define mes(a,x,s)  memset(a,x,(s)*sizeof a[0])#define mem(a,x)  memset(a,x,sizeof a)#define ysk(x)  (1<<(x))typedef long long ll;typedef pair<int, int> pii;const int INF =0x3f3f3f3f;const int maxn=500    ;string s[maxn+5];int n,T;bool vis[maxn+10];int main(){    std::ios::sync_with_stdio(false);   int T,kase=0;cin>>T;   while(T--)   {       cin>>n;       for1(i,n)  cin>>s[i];       int ans=-1;       mes(vis,0,n+1);       for(int i=1;i<=n;i++)       {          for1(j,i-1) if(!vis[j])          {              if(s[i].find(s[j])==string::npos)              {                ans=i;                break;              }              else              {                  vis[j]=1;              }          }       }       printf("Case #%d: %d\n",++kase,ans);   }   return 0;}

2.strstr()

#include<cstdio>#include<string>#include<cstring>#include<iostream>#include<cmath>#include<algorithm>#include<vector>using namespace std;#define all(x) (x).begin(), (x).end()#define for0(a, n) for (int (a) = 0; (a) < (n); (a)++)#define for1(a, n) for (int (a) = 1; (a) <= (n); (a)++)#define mes(a,x,s)  memset(a,x,(s)*sizeof a[0])#define mem(a,x)  memset(a,x,sizeof a)#define ysk(x)  (1<<(x))typedef long long ll;typedef pair<int, int> pii;const int INF =0x3f3f3f3f;const int maxn=500    ;char s[maxn+5][2010];int n;bool vis[maxn+10];int main(){   int T,kase=0;scanf("%d",&T);   while(T--)   {       scanf("%d",&n);       for1(i,n)  scanf("%s",s[i]);       int ans=-1;       mes(vis,0,n+1);       for(int i=1;i<=n;i++)       {          for1(j,i-1) if(!vis[j])          {              if(!strstr(s[i],s[j]))              {                ans=i;                break;              }              else              {                  vis[j]=1;              }          }       }       printf("Case #%d: %d\n",++kase,ans);   }   return 0;}

3.KMP

#include<cstdio>#include<string>#include<cstring>#include<iostream>#include<cmath>#include<algorithm>#include<vector>using namespace std;#define all(x) (x).begin(), (x).end()#define for0(a, n) for (int (a) = 0; (a) < (n); (a)++)#define for1(a, n) for (int (a) = 1; (a) <= (n); (a)++)#define mes(a,x,s)  memset(a,x,(s)*sizeof a[0])#define mem(a,x)  memset(a,x,sizeof a)#define ysk(x)  (1<<(x))typedef long long ll;typedef pair<int, int> pii;const int INF =0x3f3f3f3f;const int maxn=500    ;char s[maxn+5][2010];int n,nex[2010];bool vis[maxn+10];void getnex(char *P){    nex[0]=nex[1]=0;    int len=strlen(P);    for(int i=1;i<len;i++)    {        int j=nex[i];        while(j&&P[j]!=P[i])  j=nex[j];        nex[i+1]= P[j]==P[i]?j+1:0;    }}bool match(char* T,char *P){    int j=0;    int L=strlen(T),len=strlen(P);    for(int i=0;i<L;i++)    {        while(j&&T[i]!=P[j]) j=nex[j];        if(T[i]==P[j])  j++;        if(j==len)  return true;    }    return false;}bool find(char *T,char *P){      getnex(P);      return match(T,P);}int main(){   int T,kase=0;scanf("%d",&T);   while(T--)   {       scanf("%d",&n);       for1(i,n)  scanf("%s",s[i]);       int ans=-1;       mes(vis,0,n+1);       for(int i=1;i<=n;i++)       {          for1(j,i-1) if(!vis[j])          {              if(!find(s[i],s[j]))              {                ans=i;                break;              }              else              {                  vis[j]=1;              }          }       }       printf("Case #%d: %d\n",++kase,ans);   }   return 0;}