HDU-4749 Parade Show KMP算法 | DP

　　题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=4749

　　题意：给两个串S和P，求S串中存在多少个与P串的大小关系一样的串。

　　因为数字的范围是1<=k<=25之间，所以可以暴力的求25*25次KMP。当然完全没有必要这样做，在KMP的时候记录各个数的所表示的数就可以了，只需要求一遍KMP，复杂度降为O(25*n)。

 //STATUS:C++_AC_125MS_1596KB
 #include <functional>
 #include <algorithm>
 #include <iostream>
 //#include <ext/rope>
 #include <fstream>
 #include <sstream>
 #include <iomanip>
 #include <numeric>
 #include <cstring>
 #include <cassert>
 #include <cstdio>
 #include <string>
 #include <vector>
 #include <bitset>
 #include <queue>
 #include <stack>
 #include <cmath>
 #include <ctime>
 #include <list>
 #include <set>
 #include <map>
 using namespace std;
 #pragma comment(linker,"/STACK:102400000,102400000")
 //using namespace __gnu_cxx;
 //define
 #define pii pair<int,int>
 #define mem(a,b) memset(a,b,sizeof(a))
 #define lson l,mid,rt<<1
 #define rson mid+1,r,rt<<1|1
 #define PI acos(-1.0)
 //typedef
 typedef __int64 LL;
 typedef unsigned __int64 ULL;
 //const
 const int N=;
 const int INF=0x3f3f3f3f;
 const int MOD=,STA=;
 const LL LNF=1LL<<;
 const double EPS=1e-;
 const double OO=1e60;
 const int dx[]={-,,,};
 const int dy[]={,,,-};
 const int day[]={,,,,,,,,,,,,};
 //Daily Use ...
 inline int sign(double x){return (x>EPS)-(x<-EPS);}
 template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;}
 template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}
 template<class T> inline T lcm(T a,T b,T d){return a/d*b;}
 template<class T> inline T Min(T a,T b){return a<b?a:b;}
 template<class T> inline T Max(T a,T b){return a>b?a:b;}
 template<class T> inline T Min(T a,T b,T c){return min(min(a, b),c);}
 template<class T> inline T Max(T a,T b,T c){return max(max(a, b),c);}
 template<class T> inline T Min(T a,T b,T c,T d){return min(min(a, b),min(c,d));}
 template<class T> inline T Max(T a,T b,T c,T d){return max(max(a, b),max(c,d));}
 //End

 int next[N],s[N],p[N],w[],f[];
 int T,n,m,k;

 void getnext(int *s,int len)
 {
     int j=,k=-;
     next[]=-;
     while(j<len){
         if(k==- || s[k]==s[j])
             next[++j]=++k;
         else k=next[k];
     }
 }

 int solve()
 {
     int i,j,ret=,x,la=-;
     for(i=j=;i<n;i++){
         while(){
             for(x=;x<=k;x++){
                 if(f[x]>=j){w[x]=-;continue;}
                 w[x]=p[i-j+f[x]];
             }
             if((j==-||w[s[j]]==-) || p[i]==w[s[j]])break;
             j=next[j];
         }
         j++;
         if(j==m && i>=la){la=i+m;ret++;}
     }
     return ret;
 }

 int main(){
  //   freopen("in.txt","r",stdin);
     int i,j,ans;
     while(~scanf("%d%d%d",&n,&m,&k))
     {
         for(i=;i<n;i++){
             scanf("%d",&p[i]);
         }
         mem(f,-);
         for(i=;i<m;i++){
             scanf("%d",&s[i]);
             if(f[s[i]]==-)f[s[i]]=i;
         }
         p[n]=;s[m]=;

         getnext(s,m);
         ans=solve();

         printf("%d\n",ans);
     }
     return ;
 }