hdu 1711Number Sequence (KMP入门,子串第一次出现的位置)

Number Sequence

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 15149    Accepted Submission(s): 6644

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

题意：给两个串a,b;求b在a中首次出现的位置，若没有則输出-1;

#include<iostream>
#include<vector>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <math.h>
#include<algorithm>
#define ll long long
#define eps 1e-8
using namespace std;

int nexts[1000050];
int  str[1000050],b[10050];

void pre_nexts(int m)
{
    memset(nexts,0,sizeof(nexts));
    int j = 0,k = -1;
    nexts[0] = -1;
    while(j < m)
    {
        if(k == -1 || b[j] == b[k])  nexts[++j] = ++k;
        else k = nexts[k];
    }
}
int KMP(int n,int m)
{
    int i,j = 0;
    for(i = 0; i < n; )
    {
        if(str[i] == b[j])
        {
            if(j == m-1) return i - (m - 1)+1;//匹配成功，返回起始坐标
            i++,j++;
        }
        else
        {
            j = nexts[j];
            if(j == -1) { i++,j = 0;}//否则重新匹配
        }
    }
    return -1;
}
int main(void)
{
    int t,i,m,n;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d %d",&n,&m);
        for(i = 0; i < n; i++)
            scanf("%d",&str[i]);
        for(i = 0; i < m; i++)
            scanf("%d",&b[i]);
        if(n < m) { printf("-1\n"); continue;}//细节
        pre_nexts(m);
        printf("%d\n",KMP(n,m));
    }
    return 0;
}