hdoj1711(KMP模板题)Number Sequence

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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

给出一个主序列，和一个匹配序列，如果能够匹配，则输出匹配序列第一个数在主序列中的位置.

这道题是kmp的基础题.

kmp算法链接：http://www.cnblogs.com/c-cloud/p/3224788.html

代码如下：

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;int nextt[10010],T[10010];int S[1000100];int n,m,ans;void get_next(){int i=1,j=0;nextt[1]=0;//next 数组的前两项其实是可以直接算出来的，next[1]=0,next[1]=1; while(i<m){if(j==0||T[i]==T[j])//T[i]表示后缀单个字符，T[j]表示前缀单个字符 {++i;++j;nextt[i]=j;}elsej=nextt[j];//若字符不同，则j回溯 }}void kmp(){int i=1,j=1;while(i<=n&&j<=m){if(j==0||S[i]==T[j]){++i,++j;}elsej=nextt[j];}if(j>m)printf("%d\n",i-m);else   printf("-1\n");}int main(){int t;scanf("%d",&t);while(t--){scanf("%d%d",&n,&m);int i,j;for(i=1;i<=n;i++)scanf("%d",&S[i]);for(i=1;i<=m;i++)scanf("%d",&T[i]);S[0]=T[0]=-1;get_next();kmp();}}

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