【HDU 1711】Number Sequence（kmp）

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Number Sequence

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 29395    Accepted Submission(s): 12363

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

 

Recommend

lcy

题意：找到s2在s1中第一次出现的位置，没有找到输出-1

思路：kmp稍微变形一下，依然可以用来处理数字字符串，只要找到s1中出现s2就直接返回下标，若在循环内没找到，在循环结束后判断s2是不是s1的后缀就行了

代码：

#include<iostream>#include<cstring>#include<stdio.h>using namespace std;const int MAX=1e6+10;int s[MAX];int t[MAX];int Next[MAX]; int n,m;int kmp(){int i=0,j=0;while(i<n){if(j==-1||s[i]==t[j]){i++;j++;if(j==m)return i-m+1;}elsej=Next[j];}return -1;}void getNext(){int i=0,j=0;Next[0]=-1;j=Next[i];while(i<m){if(j==-1||t[i]==t[i]){Next[++i]=++j;}else{j=Next[j];}}}int main(){int T;cin>>T;while(T--){cin>>n>>m;for(int i=0;i<n;i++)scanf("%d",&s[i]);for(int i=0;i<m;i++)scanf("%d",&t[i]);getNext();cout<<kmp()<<endl;}return 0;}