HDU 5918 KMP/模拟

Sequence I

Time Limit: 3000/1500 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1013    Accepted Submission(s): 393

Problem Description

Mr. Frog has two sequences a1,a2,⋯,an and b1,b2,⋯,bm and a number p. He wants to know the number of positions q such that sequence b1,b2,⋯,bm is exactly the sequence aq,aq+p,aq+2p,⋯,aq+(m−1)p where q+(m−1)p≤n and q≥1 .

 

Input

The first line contains only one integer T≤100 , which indicates the number of test cases.

Each test case contains three lines.

The first line contains three space-separated integers 1≤n≤106,1≤m≤106 and 1≤p≤106 .

The second line contains n integers a1,a2,⋯,an(1≤ai≤109) .

the third line contains m integers b1,b2,⋯,bm(1≤bi≤109) .

 

Output

For each test case, output one line “Case #x: y”, where x is the case number (starting from 1) and y is the number of valid q’s.

 

Sample Input

2

6 3 1

1 2 3 1 2 3

1 2 3

6 3 2

1 3 2 2 3 1

1 2 3

 

Sample Output

Case #1: 2

Case #2: 1

 

Source

2016中国大学生程序设计竞赛（长春）-重现赛

 

题意：

题解：

 

kmp

/******************************
code by drizzle
blog: www.cnblogs.com/hsd-/
^ ^    ^ ^
 O      O
******************************/
#include<bits/stdc++.h>
#include<map>
#include<set>
#include<cmath>
#include<queue>
#include<bitset>
#include<math.h>
#include<vector>
#include<string>
#include<stdio.h>
#include<cstring>
#include<iostream>
#include<algorithm>
#pragma comment(linker, "/STACK:102400000,102400000")
using namespace std;
#define  A first
#define B second
const int mod=1000000007;
const int MOD1=1000000007;
const int MOD2=1000000009;
const double EPS=0.00000001;
//typedef long long ll;
typedef __int64 ll;
const ll MOD=1000000007;
const int INF=1000000010;
const ll MAX=1ll<<55;
const double eps=1e-5;
const double inf=~0u>>1;
const double pi=acos(-1.0);
typedef double db;
typedef unsigned int uint;
typedef unsigned long long ull;
int f[2000100];
void get(int *p,int m)
{
    int j=0;
    f[0]=f[1]=0;
    for(int i=1;i<m;i++)
    {
        j=f[i];
        while(j&&p[j]!=p[i]) j=f[j];
        if(p[i]==p[j]) f[i+1]=j+1;
        else f[i+1]=0;
    }
}
int kmp(int *s,int *p,int n,int m)
{
    int num=0;
    int j=0;
    for(int i=0;i<n;i++)
    {
        while(j&&p[j]!=s[i]) j=f[j];
        if(s[i]==p[j]) j++;
        if(j==m) num++;
    }
    return num;
}
int s[2000100],p[2000100],t[2000100];
int n,m,q;
int main()
{
    int T;
    scanf("%d",&T);
    for(int k=1;k<=T;k++)
    {
        scanf("%d %d %d",&n,&m,&q);
        memset(t,0,sizeof(t));
        memset(p,0,sizeof(p));
        for(int i=0;i<n;i++)
            scanf("%d",&t[i]);
        for(int i=0;i<m;i++)
            scanf("%d",&p[i]);
        get(p,m);
        int ans=0;
        for(int i=0;i<q;i++)
        {
            int num=0;
            for(int j=i;j<n&&i+(m-1)*q<n;j+=q)
                s[num++]=t[j];
            ans+=kmp(s,p,num,m);
        }
        printf("Case #%d: %d
",k,ans);
    }
    return 0;
}
/*
KMP 处理
*

模拟

/******************************
code by drizzle
blog: www.cnblogs.com/hsd-/
^ ^    ^ ^
 O      O
******************************/
#include<bits/stdc++.h>
#include<map>
#include<set>
#include<cmath>
#include<queue>
#include<bitset>
#include<math.h>
#include<vector>
#include<string>
#include<stdio.h>
#include<cstring>
#include<iostream>
#include<algorithm>
#pragma comment(linker, "/STACK:102400000,102400000")
using namespace std;
#define  A first
#define B second
const int mod=1000000007;
const int MOD1=1000000007;
const int MOD2=1000000009;
const double EPS=0.00000001;
//typedef long long ll;
typedef __int64 ll;
const ll MOD=1000000007;
const int INF=1000000010;
const ll MAX=1ll<<55;
const double eps=1e-5;
const double inf=~0u>>1;
const double pi=acos(-1.0);
typedef double db;
typedef unsigned int uint;
typedef unsigned long long ull;
int t;
int a[1000006];
int b[1000006];
int d[1000006];
map<int,int> mp;
vector<int > ve[1000006];
int n,m,p;
int main()
{
    while(scanf("%d",&t)!=EOF)
    {
        for(int l=1; l<=t; l++)
        {
            mp.clear();
            scanf("%d %d %d",&n,&m,&p);
            for(int i=1; i<=n; i++)
                scanf("%d",&a[i]);
            int jishu=1;
            for(int i=1; i<=m; i++ )
            {
                ve[i].clear();
                scanf("%d",&b[i]);
                if(mp[b[i]]==0)
                {
                    mp[b[i]]=jishu;
                    d[jishu]=i;
                    jishu++;
                }
            }
            int minx=10000000;
            int what=0;
            for(int i=1; i<=n; i++)
            {
                if(mp[a[i]])
                {
                    ve[mp[a[i]]].push_back(i);
                }
            }
            for(int i=1; i<jishu; i++)
            {
                if(minx>ve[i].size())
                {
                    minx=ve[i].size();
                    what=i;
                }
            }
            int sum=0;
            for(int i=0; i<ve[what].size(); i++)
            {
                int st=ve[what][i]-(d[mp[a[ve[what][i]]]]-1)*p;
                int ed=ve[what][i]+(m-d[mp[a[ve[what][i]]]])*p;
                int zha=1;
                int flag=0;
                if(st<1||ed>n)
                    flag=1;
                if(flag==0)
                {
                    for(int j=st; j<=ed; j+=p)
                    {
                        if(a[j]!=b[zha++])
                        {
                            flag=1;
                            break;
                        }
                    }
                }
                if(flag==0)
                    sum++;
            }
            printf("Case #%d: %d
",l,sum);
        }
    }
    return 0;
}