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 }
