HDU 5918 Sequence I KMP

Sequence I



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,,bmis exactly the sequence aq,aq+p,aq+2p,,aq+(m1)p where q+(m1)pn and q1.
 

 

Input
 
The first line contains only one integer T100, which indicates the number of test cases.

Each test case contains three lines.

The first line contains three space-separated integers 1n106,1m106 and 1p106.

The second line contains n integers a1,a2,,an(1ai109).

the third line contains m integers b1,b2,,bm(1bi109).
 

 

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
 

题意:

  给你a,b两个序列

  和一个p ,求有多少个 q恰好满足 b1,b2,b3....bm 就是 a[q],a[q+p],a[q+2p]......a[q+(m-1)p];

题解:

  将a序列,每隔p位置分成一组,这样最多有p组,个数和是n

  将每组和b序列跑KMP计算答案

#include<bits/stdc++.h>
using namespace std;
#pragma comment(linker, "/STACK:102400000,102400000")
#define ls i<<1
#define rs ls | 1
#define mid ((ll+rr)>>1)
#define pii pair<int,int>
#define MP make_pair
typedef long long LL;
const long long INF = 1e18;
const double Pi = acos(-1.0);
const int N = 1e6+10, M = 1e6, mod = 1e9+7, inf = 2e9;


int T,n,m,p,s[N],t[N],ans = 0;
vector<int > P[N];
int nex[N];
int main()
{
    int cas = 1;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d%d",&n,&m,&p);
        for(int i = 1; i <= n; ++i) scanf("%d",&t[i]);
        for(int i = 1; i <= m; ++i) scanf("%d",&s[i]);
        for(int i = 0; i < p; ++i) P[i].clear();
        memset(nex,0,sizeof(nex));
        ans = 0;
        int k = 0;
        for(int i=2; i<=m; i++)
        {
            while(k>0&&s[k+1]!=s[i]) k = nex[k];
            if(s[k+1]==s[i])k++;
            nex[i] = k;
        }
        if(p == 1)
        {
            k = 0;
            for(int i=1; i<=n; i++)
            {
                while(k>0&&s[k+1]!=t[i]) k = nex[k];
                if(s[k+1]==t[i]) k++;
                if(k==m) {
                    k = nex[k];
                    ans++;
                }
            }
            printf("Case #%d: ",cas++);
            printf("%d\n",ans);
        }
        else {
            for(int i = 1; i <= n; ++i)
                P[i % p].push_back(t[i]);
            for(int i = 0; i < p; ++i) {
                k = 0;
                for(int j = 0; j < P[i].size(); ++j) {
                    while(k>0&&s[k+1]!=P[i][j]) k = nex[k];
                    if(s[k+1]==P[i][j]) k++;
                    if(k==m) {
                        k = nex[k];
                        ans++;
                    }
                }
            }
            printf("Case #%d: ",cas++);
            printf("%d\n",ans);

        }
    }
}

 

posted @ 2016-10-04 20:29  meekyan  阅读(280)  评论(0编辑  收藏  举报