HDOJ---1711 Number Sequence[KMP模版]

Number Sequence

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 6147    Accepted Submission(s): 2758

Problem Description

Given two sequences of numbers : a[1], a[2], ...... , a[N], and b[1], b[2], ...... , b[M] (1 <= M <= 10000, 1 <= N <= 1000000). Your task is to find a number K which make a[K] = b[1], a[K + 1] = b[2], ...... , a[K + M - 1] = b[M]. If there are more than one K exist, output the smallest one.

 

 

Input

The first line of input is a number T which indicate the number of cases. Each case contains three lines. The first line is two numbers N and M (1 <= M <= 10000, 1 <= N <= 1000000). The second line contains N integers which indicate a[1], a[2], ...... , a[N]. The third line contains M integers which indicate b[1], b[2], ...... , b[M]. All integers are in the range of [-1000000, 1000000].

 

 

Output

For each test case, you should output one line which only contain K described above. If no such K exists, output -1 instead.

 

 

Sample Input

2 13 5 1 2 1 2 3 1 2 3 1 3 2 1 2 1 2 3 1 3 13 5 1 2 1 2 3 1 2 3 1 3 2 1 2 1 2 3 2 1

 

 

Sample Output

6 -1

 

 

Source

HDU 2007-Spring Programming Contest

 

 

Recommend

lcy

 

 

 

题意：求完全匹配时匹配串的位置

 

注意数组大小

 

code:

#include<iostream>
using namespace std;

#define MAXN  10000000

int p[MAXN];
int s[MAXN];
int Next[MAXN];
int n,m;

void getNext()
{
    int j,k;
    j=0;
    k=-1;
    Next[0]=-1;
    while(j<m)
    {
        if(k==-1||p[j]==p[k])
        {
            j++;
            k++;
            Next[j]=k;
        }
        else
            k=Next[k];
    }
}

int kmp()
{
    int i,j;
    i=j=0;
    getNext();
    while(i<n)
    {
        if(j==-1||s[i]==p[j])
        {
            i++;
            j++;
        }
        else
        {
            j=Next[j];
        }
        if(j==m)
            return i;
    }
    return -1;
}


int main()
{
    int t;
    int i;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&n,&m);
        for(i=0;i<n;i++)
            scanf("%d",&s[i]);
        for(i=0;i<m;i++)
            scanf("%d",&p[i]);
        if(kmp()==-1)
            printf("-1\n");
        else
            printf("%d\n",kmp()-m+1);
    }
    return 0;
}