【字符匹配专题】--1001 kmp

Number Sequence

Time Limit : 10000/5000ms (Java/Other)   Memory Limit : 32768/32768K (Java/Other)

Total Submission(s) : 10   Accepted Submission(s) : 5

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

 

 

 

这个是一个简单的KMP模板的运用，只不过将字符串变成了数组，因此，稍作调整就可A过去，但第一次提交却TLE了....看了看，就把c++中的cin改成scanf...AC了，╮(╯▽╰)╭，习惯使用C++输入输出的伤不起啊...

 

#include <iostream>using namespace std;int a[1000005],b[10005],x,y;int next[10005];void get_next( ){int i,j;i=1;next[1]=0;j=0;while(i<y){if(j==0||b[i-1]==b[j-1]){i++;j++;if(b[i-1]!=b[j-1])next[i]=j;elsenext[i]=next[j];}elsej=next[j];}}int kmp(){get_next();int i,j;//int lens,lent;i=1;j=1;while(i<=x&&j<=y){if(j==0||a[i-1]==b[j-1]){i++;j++;}elsej=next[j];}if(j>y)return i-y;elsereturn -1;}int main(){int n;cin>>n;while(n--){//cin>>x>>y;scanf("%d%d",&x,&y);for(int i=0;i<x;i++)    //cin>>a[i];    scanf("%d",&a[i]);        for(int i=0;i<y;i++)            //cin>>b[i];            scanf("%d",&b[i]);                if(x<y)        {         cout<<-1<<endl;         return 0;        }                        int ans=kmp();        cout<<ans<<endl;        }return 0;}