hdu 1711 Number Sequence KMP 基础题

Number Sequence

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 11691    Accepted Submission(s): 5336

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

 

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题意：给两个数组a和b，求b在a中出现的的第一个位置，若没有则输出-1，仅仅是数组变成了整型数组，方法与字符数组全然一样。

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <string>
#include <map>
#include <stack>
#include <vector>
#include <set>
#include <queue>
#pragma comment (linker,"/STACK:102400000,102400000")
#define maxn 1000005
#define MAXN 2005
#define mod 1000000009
#define INF 0x3f3f3f3f
#define pi acos(-1.0)
#define eps 1e-6
typedef long long ll;
using namespace std;

int s[maxn],p[maxn],nextval[maxn];
int N,M;

void get_nextval()
{
    int i=0,j=-1;
    nextval[0]=-1;
    while (i<M)
    {
        if (j==-1||p[i]==p[j])
        {
            i++;
            j++;
            if (p[i]!=p[j])
                nextval[i]=j;
            else
                nextval[i]=nextval[j];
        }
        else
            j=nextval[j];
    }
}

int KMP()
{
    int i=0,j=0;
    while (i<N&&j<M)
    {
        if (j==-1||s[i]==p[j])
        {
            i++;
            j++;
        }
        else
            j=nextval[j];
    }
    if (j==M)
        return i-M+1;
    else
        return -1;
}

int main()
{
    int cas;
    scanf("%d",&cas);
    while (cas--)
    {
        scanf("%d%d",&N,&M);
        for (int i=0;i<N;i++)
            scanf("%d",&s[i]);
        for (int i=0;i<M;i++)
            scanf("%d",&p[i]);
        get_nextval();
        int ans=KMP();
        printf("%d\n",ans);
    }
    return 0;
}
/*
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
*/