HDU 1711 Number Sequence

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本文介绍了一道经典的KMP字符串匹配题目，通过给出的代码实现了解决方案，旨在寻找两个整数序列中子串的起始位置。文章详细解释了KMP算法的原理及其实现细节。

题目链接： http://hdu.hustoj.com/showproblem.php?pid=1711

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
 

题意：找出第一个子串的位置，没有输出-1

思路：KMP裸题，用index函数找一下位置

代码：

#include<cstdio>
#include<cmath>
#include<cstring>
#include<string>
#include<cstdlib>
#include<algorithm>
#include<iostream>
#include<queue>
#include<stack>
#include<map>

using namespace std;

#define FOU(i,x,y) for(int i=x;i<=y;i++)
#define FOD(i,x,y) for(int i=x;i>=y;i--)
#define MEM(a,val) memset(a,val,sizeof(a))
#define PI acos(-1.0)

const double EXP = 1e-9;
typedef long long ll;
typedef unsigned long long ull;
const int INF = 0x3f3f3f3f;
const ll MINF = 0x3f3f3f3f3f3f3f3f;
const int mod = 1e9+7;
const int N = 1e6+5;

int n,m;
int a[N];
int Next[10005];
int mo[10005];

void Get_nextval()
{
    int i,j;
    j = Next[0] = -1;
    i=0;
    while(i<m)
    {
        while(j!=-1&&mo[i]!=mo[j])
            j = Next[j];
        if(mo[++i]==mo[++j])
            Next[i] = Next[j];
        else
            Next[i] = j;
    }
}

void Get_next()
{
    int j=-1;
    Next[0]=-1;
    int i=0;
    while(i<m)
    {
        while(j!=-1&&mo[i]!=mo[j])
            j=Next[j];
        Next[++i]=++j;
    }
}

int index()
{
    int i=0,j=0;
    while(i<n&&j<m)
    {
        if(a[i]==mo[j]||j==-1)
            i++,j++;
        else
            j=Next[j];
    }
    if(j==m)
        return i-m+1;
    return -1;
}

int main()
{
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    std::ios::sync_with_stdio(false);
    int t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&n,&m);
        for(int i=0;i<n;i++)
            scanf("%d",&a[i]);
        for(int i=0;i<m;i++)
            scanf("%d",&mo[i]);
        Get_nextval();
        int ans = index();
        printf("%d\n",ans);
    }
    return 0;
}