【HDU-1711】解题报告（KMP模板题，字符串hash）

原始题目

Number Sequence

  • Time Limit: 10000/5000 MS (Java/Others)
  • Memory Limit: 32768/32768 K (Java/Others)
  • Total Submission(s): 39142
  • Accepted Submission(s): 16165

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

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题目大意

  • 给定一个数字串和一个模式串，输出最早匹配的位置，没有输出-1。

解题思路

  • KMP把字符匹配改成int即可。

解题代码

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//#include <bits/stdc++.h>
#include <cstdio> 
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <iomanip>
#include <string>
#include <map>
#include <set>
#include <algorithm>
#include <queue>
#include <vector>
#define pb push_back
#define mp make_pair
using namespace std;
typedef long long LL;
typedef pair<int,int> PII;
#define PI acos((double)-1)
#define E exp(double(1))
#define K 1000000+9
#define rep(i,a,n) for(int i=a;i<n;i++)
#define per(i,a,n) for(int i=n-1;i>=a;i--)
#define ms(x,a) memset((x),(a),sizeof(a))
int nt[10006];
int a[K],b[10005],n,m;
//参数为模板串和next数组
//字符串均从下标0开始
void kmp_next(int *T,int *nt)
{
    nt[0]=0;
    for(int i=1,j=0;i<m;i++)
    {
        while(j&&T[i]!=T[j])j=nt[j-1];
        if(T[i]==T[j])j++;
        nt[i]=j;
    }
}
int kmp(int *S,int *T,int *nt)
{
    kmp_next(T,nt);
    int ans=0;
    //n //m
    for(int i=0,j=0;i<n;i++)
    {
        while(j&&S[i]!=T[j])j=nt[j-1];
        if(S[i]==T[j])j++;
        if(j==m){
//        	ans++;
			return i-m+2;
		}
           
    }
    return -1;
}
int main(void)
{
	ios::sync_with_stdio(false);
    int t;cin>>t;
    while(t--)
    {
    	cin>>n>>m;
    	rep(i,0,n) cin>>a[i];
    	rep(i,0,m) cin>>b[i];
    	cout<< kmp(a,b,nt)<<endl;
    }

    return 0;
}
//1 2 1 2 3 1 2 3 1 3  2 1 2 3 1 2 3

收获与反思

  • 单模匹配考虑KMP，熟悉模板和算法。