String match
Description [title]
For the size of a character set C character string PP , may be any two characters in the p position interchanging, e.g. p = 12321 , exchange 1 , 21 , 2 to obtain 21312 , exchange 1 , 4 to obtain 42,324 , can exchange any number of times. If the exchange of p into a string of q , to the q and p are matched.
Given the size of two character sets C string s , t , to obtain s the number of contiguous substrings have t match.
[Enter]
The first line of two integers T , C , respectively, and the number of data sets the size of characters, characters . 1 ~ C integers represented.
For each set of data: The first line of two integers n- , m , respectively S , T length.
The second line n positive integer S .
The third row m positive integer T .
[Output]
For each test, two output lines comprising:
The first line of a positive integer k , represents s has k consecutive sub strings t match.
The second row from small to large outputs k number indicating s with t match the first string of successive sub index (subscript from 1 starts).
[Sample input]
3 3
6 3
1 2 1 2 3 2
3 1 3
6 3
1 2 1 2 1 2
3 1 3
6 3
1 1 2 1 2 1
3 1 3
[Sample Output]
3
1 2 4
4
1 2 3 4
3
2 3 4
[Agreement] and the size of data
For 10% of the data, to meet the n-, m, C ≤ 1000N, m, C ≤ 1000 ;
For another 20% of the data, to meet the n-, m ≤ 105 , C ≤ 40N, m ≤ 105 , C ≤ 40 ;
For another 30% of the data, to meet the n-, m, C ≤ 105n, m, C ≤ 105 ;
For 100% data meet . 1 ≤ n-, m, C ≤ 106 , T = 31 is ≤ n-, m, C ≤ 106 , T =. 3 .
【analysis】
This is actually a KMP title
The difficulty is how to exchange the title character
We can start an array l [1 ~ c]
L [x] on showing an x position occurring
a [i] representing the character s [i] a distance from appearing on a same character
b [i] represents the character t [i] a distance from appearing on a same character
Then there KMP
Difficulty is determining how to s [i] (t [i ]) than in the pattern string is
Well this is actually very simple
Directly determining a x whether the distance is greater than the occurrence j not on the list
AC [ Code ]
#include<cstdio>
#include<cstring>
#include<algorithm>
#define N (1000000+2)
#define C (1000000+2)
using namespace std;
int p[N];
int a[N],b[N];
int l[C];
int ans[N];
template<typename T>inline void read(T& x){
char temp=getchar();bool u=0;
for(x=0;temp<'0'||temp>'9';u=temp=='-',temp=getchar());
for(;temp>='0'&&temp<='9';x=x*10+temp-'0',temp=getchar());
if(u)x=-x;
return ;
}
inline int jdg(int x,int l){return x<l?x:0;}
inline bool eq(int x,int y,int l){return jdg(x,l)==jdg(y,l);}
void work(){
register int i,j,k;
register int n,m;
read(n);
read(m);
memset(a,0,sizeof a);
memset(b,0,sizeof b);
memset(p,0,sizeof p);
memset(l,0,sizeof l);
for(i=1;i<=n;i++){
read(k);
a[i]=i-l[k];
l[k]=i;
}
memset(l,0,sizeof l);
for(i=1;i<=m;i++){
read(k);
b[i]=i-l[k];
l[k]=i;
}
for(i=1,j=0;i<m;i++){
while(j&&!eq(b[i+1],b[j+1],j+1))
j=p[j];
if(eq(b[i+1],b[j+1],j+1))
j++;
p[i+1]=j;
}
for(k=i=j=0;i<n;i++){
while(j&&!eq(a[i+1],b[j+1],j+1))
j=p[j];
if(eq(a[i+1],b[j+1],j+1))
j++;
if(j==m){
ans[++k]=i+2-m;
j=p[j];
}
}
printf("%d\n",k);
for(i=1;i<=k;i++)
printf("%d ",ans[i]);
putchar('\n');
return ;
}
int main(){
register int t,c,i;
read(t);
read(c);
for(i=1;i<=t;i++)
work();
return 0;
}