hdu 1711 Number Sequence （kmp）

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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

213 51 2 1 2 3 1 2 3 1 3 2 1 21 2 3 1 313 51 2 1 2 3 1 2 3 1 3 2 1 21 2 3 2 1

Sample Output

6-1

Source

HDU 2007-Spring Programming Contest

题意：给出两个长度分别为n和m的数组a和b，求出a完全匹配b的首个序列，并输出a对应的第一个位置；

思路：cmp模板；

代码：

#include<cstdio>#include<algorithm>#include<cstring>using namespace std;const int M=10005;const int N=1000005;int nex[M];int a[N],b[M];int n,m;void getnext(){    int i=0,j=-1;    nex[0]=-1;    while(b[i])    {        if(j==-1||b[i]==b[j])            nex[++i]=++j;        else            j=nex[j];    }}int cmp(){    int vis=0;    for(int i=0;i<n;i++)    {        if(a[i]==b[vis])        {            vis++;            if(vis==m)                return i-m+2;            continue;        }        if(vis==0) //只匹配到第一个，所以往下继续；            continue;        vis=nex[vis];        i--;   //因为不匹配，我们还要继续匹配之前已经匹配的；    }    return -1;}int main(){    int t;    scanf("%d",&t);    while(t--)    {        scanf("%d%d",&n,&m);        memset(nex,0,sizeof(nex));        for(int i=0;i<n;i++)            scanf("%d",&a[i]);        for(int i=0;i<m;i++)            scanf("%d",&b[i]);        getnext();        printf("%d\n",cmp());    }}

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