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)。

  1 //STATUS:C++_AC_125MS_1596KB

  2 #include <functional>

  3 #include <algorithm>

  4 #include <iostream>

  5 //#include <ext/rope>

  6 #include <fstream>

  7 #include <sstream>

  8 #include <iomanip>

  9 #include <numeric>

 10 #include <cstring>

 11 #include <cassert>

 12 #include <cstdio>

 13 #include <string>

 14 #include <vector>

 15 #include <bitset>

 16 #include <queue>

 17 #include <stack>

 18 #include <cmath>

 19 #include <ctime>

 20 #include <list>

 21 #include <set>

 22 #include <map>

 23 using namespace std;

 24 #pragma comment(linker,"/STACK:102400000,102400000")

 25 //using namespace __gnu_cxx;

 26 //define

 27 #define pii pair<int,int>

 28 #define mem(a,b) memset(a,b,sizeof(a))

 29 #define lson l,mid,rt<<1

 30 #define rson mid+1,r,rt<<1|1

 31 #define PI acos(-1.0)

 32 //typedef

 33 typedef __int64 LL;

 34 typedef unsigned __int64 ULL;

 35 //const

 36 const int N=100010;

 37 const int INF=0x3f3f3f3f;

 38 const int MOD=9973,STA=8000010;

 39 const LL LNF=1LL<<60;

 40 const double EPS=1e-6;

 41 const double OO=1e60;

 42 const int dx[4]={-1,0,1,0};

 43 const int dy[4]={0,1,0,-1};

 44 const int day[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};

 45 //Daily Use ...

 46 inline int sign(double x){return (x>EPS)-(x<-EPS);}

 47 template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;}

 48 template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}

 49 template<class T> inline T lcm(T a,T b,T d){return a/d*b;}

 50 template<class T> inline T Min(T a,T b){return a<b?a:b;}

 51 template<class T> inline T Max(T a,T b){return a>b?a:b;}

 52 template<class T> inline T Min(T a,T b,T c){return min(min(a, b),c);}

 53 template<class T> inline T Max(T a,T b,T c){return max(max(a, b),c);}

 54 template<class T> inline T Min(T a,T b,T c,T d){return min(min(a, b),min(c,d));}

 55 template<class T> inline T Max(T a,T b,T c,T d){return max(max(a, b),max(c,d));}

 56 //End

 57 

 58 int next[N],s[N],p[N],w[30],f[30];

 59 int T,n,m,k;

 60 

 61 void getnext(int *s,int len)

 62 {

 63     int j=0,k=-1;

 64     next[0]=-1;

 65     while(j<len){

 66         if(k==-1 || s[k]==s[j])

 67             next[++j]=++k;

 68         else k=next[k];

 69     }

 70 }

 71 

 72 int solve()

 73 {

 74     int i,j,ret=0,x,la=-1;

 75     for(i=j=0;i<n;i++){

 76         while(1){

 77             for(x=1;x<=k;x++){

 78                 if(f[x]>=j){w[x]=-1;continue;}

 79                 w[x]=p[i-j+f[x]];

 80             }

 81             if((j==-1||w[s[j]]==-1) || p[i]==w[s[j]])break;

 82             j=next[j];

 83         }

 84         j++;

 85         if(j==m && i>=la){la=i+m;ret++;}

 86     }

 87     return ret;

 88 }

 89 

 90 int main(){

 91  //   freopen("in.txt","r",stdin);

 92     int i,j,ans;

 93     while(~scanf("%d%d%d",&n,&m,&k))

 94     {

 95         for(i=0;i<n;i++){

 96             scanf("%d",&p[i]);

 97         }

 98         mem(f,-1);

 99         for(i=0;i<m;i++){

100             scanf("%d",&s[i]);

101             if(f[s[i]]==-1)f[s[i]]=i;

102         }

103         p[n]=0;s[m]=0;

104 

105         getnext(s,m);

106         ans=solve();

107 

108         printf("%d\n",ans);

109     }

110     return 0;

111 }