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=100010;
const int INF=0x3f3f3f3f;
const int MOD=9973,STA=8000010;
const LL LNF=1LL<<60;
const double EPS=1e-6;
const double OO=1e60;
const int dx[4]={-1,0,1,0};
const int dy[4]={0,1,0,-1};
const int day[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};
//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[30],f[30];
int T,n,m,k;

void getnext(int *s,int len)
{
    int j=0,k=-1;
    next[0]=-1;
    while(j<len){
        if(k==-1 || s[k]==s[j])
            next[++j]=++k;
        else k=next[k];
    }
}

int solve()
{
    int i,j,ret=0,x,la=-1;
    for(i=j=0;i<n;i++){
        while(1){
            for(x=1;x<=k;x++){
                if(f[x]>=j){w[x]=-1;continue;}
                w[x]=p[i-j+f[x]];
            }
            if((j==-1||w[s[j]]==-1) || 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=0;i<n;i++){
            scanf("%d",&p[i]);
        }
        mem(f,-1);
        for(i=0;i<m;i++){
            scanf("%d",&s[i]);
            if(f[s[i]]==-1)f[s[i]]=i;
        }
        p[n]=0;s[m]=0;

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

        printf("%d
",ans);
    }
    return 0;
}