HDU 1711Number Sequence KMP模板题(找模板第一次出现的位置)-CFANZ编程社区

Number Sequence Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 42763 Accepted Submission(s): 17662 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 Source HDU 2007-Spring Programming Contest Recommend lcy \| We have carefully selected several similar problems for you: 1358 3336 1686 3746 2222

#include<stdio.h>
#include<string.h>
#include<iostream>
using namespace std;
 
const int N = 1000002;
int nxt[N];
int S[N], T[N];
int slen, tlen;
 
void getNext()
{
    int j, k;
    j = 0; k = -1;
    nxt[0] = -1;
    while(j < tlen)
        if(k == -1 || T[j] == T[k])
           {
            nxt[++j] = ++k;
            if (T[j] != T[k]) //优化
                nxt[j] = k; 
           } 
        else
            k = nxt[k];
 
}
/*
返回模式串T在主串S中首次出现的位置
返回的位置是从0开始的。
*/
int KMP_Index()
{
    int i = 0, j = 0;
    getNext();
    while(i < slen && j < tlen)
    {
        if(j == -1 || S[i] == T[j])
        {
            i++; 
            j++;
        }
        else
            j = nxt[j];
    }
    if(j == tlen)
        return i - tlen;
    else
        return -1;
}
int main()
{
 
    int TT;
    int i, cc;
    cin>>TT;
    int n,m;
    while(TT--)
    {
        scanf("%d%d",&n,&m);
        for(int i=0;i<n;i++)
         scanf("%d",&S[i]);
        for(int i=0;i<m;i++)
         scanf("%d",&T[i]);
        slen = n;
        tlen = m;
        if(KMP_Index()==-1)
        cout<<KMP_Index()<<endl;
        else
        cout<<KMP_Index()+1<<endl;
        
    }
    return 0;
}