hdu4749 kmp应用

呃，从网上看的题解，然而其实有点地方还没搞懂，先放在这，以后再回来理解。

 

题目地址：http://acm.hdu.edu.cn/showproblem.php?pid=4749

题目：2013 is the 60 anniversary of Nanjing University of Science and Technology, and today happens to be the anniversary date. On this happy festival, school authority hopes that the new students to be trained for the parade show. You should plan a better solution to arrange the students by choosing some queues from them preparing the parade show. (one student only in one queue or not be chosen)
  Every student has its own number, from 1 to n. (1<=n<=10^5), and they are standing from 1 to n in the increasing order the same with their number order. According to requirement of school authority, every queue is consisted of exactly m students. Because students who stand adjacent in training are assigned consecutive number, for better arrangement, you will choose in students with in consecutive numbers. When you choose these m students, you will rearrange their numbers from 1 to m, in the same order with their initial one. 
  If we divide our students’ heights into k (1<=k<=25) level, experience says that there will exist an best viewing module, represented by an array a[]. a[i] (1<=i<=m)stands for the student’s height with number i. In fact, inside a queue, for every number pair i, j (1<=i,j<=m), if the relative bigger or smaller or equal to relationship between the height of student number i and the height of student number j is the same with that between a[i] and a[j], then the queue is well designed. Given n students’ height array x[] (1<=x[i]<=k), and the best viewing module array a[], how many well designed queues can we make at most?

 

 

Input

Multiple cases, end with EOF.
First line, 3 integers, n (1<=n<=10^5) m (1<=m<=n) k(1<=k<=25),
Second line, n students’ height array x[] (1<=x[i]<=k,1<=i<=n);
Third line, m integers, best viewing module array a[] (1<=a[i]<=k,1<=i<=m);

 

 

Output

One integer, the maximal amount of well designed queues.

 

 

Sample Input

10 5 10 2 4 2 4 2 4 2 4 2 4 1 2 1 2 1

 

 

Sample Output

1

 

#include 
#define PB push_back
#define MP make_pair
using namespace std;
typedef long long LL;
typedef pair<int,int> PII;
#define PI acos((double)-1)
#define E exp(double(1))
#define K 100000+9
int g1[30][K],g2[30][K];
int a[K],b[K],n,m,k;
int nt[K];
int check(int x,int y)
{
    int c1,c2,c3;
    c1=c2=c3=0;
    if(g2[b[x]][x-1]!=g2[b[y]][y-1]-g2[b[y]][y-x])return 0;
    for(int i=1;i1];
    for(int i=1;i1];
    for(int i=1;ix];
    return c1==c2-c3;
}

void kmp_next(void)
{
    memset(nt,0,sizeof(nt));
    for(int i=2,j=0;i<=m;i++)
    {
        while(j && !check(j+1,i))
            j=nt[j];
        if(check(j+1,i))j++;
        nt[i]=j;
    }
}
bool ok(int x,int y)
{
    int c1,c2,c3;
    c1=c2=c3=0;
    if(g2[b[x]][x-1]!=g1[a[y]][y-1]-g1[a[y]][y-x])return 0;
    for(int i=1;i1];
    for(int i=1;i1];
    for(int i=1;ix];
    return c1==c2-c3;
}
int kmp(void )
{
    int ans=0;
    for(int i=1,j=0;i<=n;i++)
    {
        while(j&&!ok(j+1,i))
            j=nt[j];
        if(ok(j+1,i))
            j++;
        if(j==m)
            ans++,j=0;
    }
    return ans;
}
int main(void)
{
    while(~scanf("%d%d%d",&n,&m,&k))
    {
        memset(g1,0,sizeof(g1));
        memset(g2,0,sizeof(g2));
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&a[i]);
            for(int j=1;j<=k;j++)g1[j][i]=g1[j][i-1];
            g1[a[i]][i]++;
        }
        for(int i=1;i<=m;i++)
        {
            scanf("%d",&b[i]);
            for(int j=1;j<=k;j++)g2[j][i]=g2[j][i-1];
            g2[b[i]][i]++;
        }
        kmp_next();
        printf("%d\n",kmp());
    }

    return 0;
}

 

转载于:https://www.cnblogs.com/weeping/p/5730978.html