HDU1711-Number Sequence

Number Sequence

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 24460    Accepted Submission(s): 10387

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

 

代码：

#include<iostream>
#include<cstring>
#include<string>
#include<cmath>
#include<cstdlib>
#include<cstdio>
using namespace std;
const int MAXN=1e6+2;
int Next[MAXN];
int t[MAXN],p[MAXN];
int tlen,plen;
void getNext() {
    int j,k;
    j=0;
    k=-1;
    Next[0]=-1;
    while(j<plen) {
        if(k==-1||p[j]==p[k]) {
            j++;
            k++;
            Next[j]=k;
        } else {
            k=Next[k];
        }
    }
}
int KMP_Index() {
    int i=0,j=0;
    getNext();
    while(i<tlen&&j<plen) {
        if(j==-1||t[i]==p[j]) {
            i++;
            j++;
        } else {
            j=Next[j];
        }
    }
    if(j==plen)
        return i-j+1;
    return -1;
}
int KMP_Count() {
    int ans=0;
    int i,j=0;
    if(tlen==1&&plen==1) {
        if(t[0]==p[0])
            return 1;
        return 0;
    }
    getNext();
    for(i=0; i<tlen; i++) {
        while(j>0&&t[i]!=p[j])
            j=Next[j];
        if(t[i]==p[j])
            j++;
        if(j==plen) {
            ans++;
            j=Next[j];
        }
    }
    return ans;
}
int main() {
    int tt;
    scanf("%d",&tt);
    while(tt--)
    {
        scanf("%d %d",&tlen,&plen);
        for(int i=0;i<tlen;i++)
        {
            scanf("%d",&t[i]);
        }
        for(int i=0;i<plen;i++)
        {
            scanf("%d",&p[i]);
        }
        printf("%d\n",KMP_Index());
        //printf("%d\n",KMP_Count());
    }
    return 0;
}