2016 ccpc(长春) Sequence I (kmp)

Sequence I

Time Limit: 3000/1500 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 958    Accepted Submission(s): 370


[align=left]Problem Description[/align]
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.
 

[align=left]Input[/align]
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).
 

[align=left]Output[/align]
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.
 

[align=left]Sample Input[/align]

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

 

[align=left]Sample Output[/align]

Case #1: 2
Case #2: 1

 

[align=left]Source[/align]
2016中国大学生程序设计竞赛（长春）-重现赛

 

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当时暴力过的，没想过kmp，太菜了
#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
#define maxx 1123456
int ne[1123456];
int s[maxx],t[maxx];
int p,sum;
void get_next(int l)
{
int i=0,j=-1;
ne[0]=-1;
while(i<l)
{
if(j==-1 || t[i]==t[j])
{
i++;j++;ne[i]=j;
}
else
{
j=ne[j];
}
}
}
void kmp(int x,int ls,int lt)
{
int i=x;
int j=0;
sum=0;
while(i<ls)
{
if(j==-1 || s[i]==t[j])
{
i+=p;j++;
}
else
{
j=ne[j];
}
if(j==lt)
{
sum++;
j=ne[j];
}
}

}
int main()
{
int n,m,i;
int ll;
int tt;
scanf("%d",&tt);
for(ll=1;ll<=tt;ll++)
{
sum=0;
scanf("%d%d%d",&n,&m,&p);
for(i=0;i<n;i++)
{
scanf("%d",&s[i]);
}
for(i=0;i<m;i++)
{
scanf("%d",&t[i]);
}
get_next(m);
int ans=0;
for(i=1;i<=n;i++)
{
printf("%d ",ne[i]);
}
printf("\n");
for(i=0;i<n&&i<p;i++) //如果i大于p，就重复的找了
{
kmp(i,n,m);
ans+=sum;
}
printf("Case #%d: %d\n",ll,ans);
}

return 0;
}