HDU 1711 Number Sequence

Number Sequence

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 16796    Accepted Submission(s): 7378

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

 

KMP水题，练习总结模版，借用kuangbin的模板，顺带做几点小的说明：代码如下：

/*************************************************************************
	> File Name: Number_Sequence.cpp
	> Author: Zhanghaoran
	> Mail: chilumanxi@xiyoulinux.org
	> Created Time: Mon 23 Nov 2015 09:16:28 PM CST
 ************************************************************************/

#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <cstdlib>

using namespace std;
typedef long long ll;
int T;
int N, M;


template <class T>
inline bool scan_d(T &ret){
    char c;
    int sgn;
    if(c = getchar(), c == EOF)
        return 0;
    while(c != '-' && (c < '0' || c > '9'))
        c = getchar();
    sgn = (c == '-') ? -1 : 1;
    ret = (c == '-') ? 0 : (c - '0');
    while(c = getchar(), c >= '0' && c <= '9')
        ret = ret * 10 + (c - '0');
    ret *= sgn;
    return 1;
}

inline void out(int x){
    if(x > 9)
        out(x / 10);
    putchar(x % 10 + '0');
}

/*
 * next[]的含义是x[i - next[i]...i - 1] = x[0...next[i - 1]]
 * next[i]为满足x[i - z...i - 1] = x[0...z - 1]
 */
void kmp_pre(ll x[], int next[]){
    int i, j;
    j = next[0] = -1;
    i = 0;
    while(i < M){
        while(j != -1 && x[i] != x[j])
            j = next[j];
        next[++ i] = ++ j;
    }
}
/*
 * kmpNext[]的含义是next'[i] = next[next[...[next[i]]]]（直到next'[i] < 0或者是x[next'[i] != x[i]]）
 * 这样的预处理更快些
 */

void pre_KMP(ll x[], int kmpNext[]){
    int i, j;
    j = kmpNext[0] = -1;
    i = 0;
    while(i < M){
        while(j != -1 && x[i] != x[j])
            j = kmpNext[j];
        if(x[++ i] == x[++ j])
            kmpNext[i] = kmpNext[j];
        else 
            kmpNext[i] = j;
    }
}

int nexti[10010];
int KMP_Count(ll x[], ll y[]){
    int i, j;
    int ans = 0;
    //kmp_pre(x, nexti);
    pre_KMP(x, nexti);
    i = j = 0;
    while(i < N){
        while(j != -1 && y[i] != x[j])
            j = nexti[j];
        i ++;
        j ++;
        if(j >= M){
            return i - M + 1;
        }
    }
    return 0;
}
ll a[1000010];
ll b[10010];
int main(void){
    cin >> T;
    while(T --){
        cin >> N >> M;
        for(int i = 0; i < N; i ++){
            scanf("%lld", &a[i]);
        }
        for(int j = 0; j < M; j ++){
            scanf("%lld", &b[j]);
        }
        int temp = KMP_Count(b, a);
        if(temp){
            out(temp);
            puts("");
        }
        else 
            puts("-1");
    }
}

当然如果有兴趣看过这篇文章全套字符串匹配算法。会发现我还记录了很多的其他的字符串匹配算法，并且看起来比KMP都要快，但是这个题目的特殊之处和KMP引以为傲的所在就是在他不但可以针对字符串，更可以针对数字串，如果对于类似Sunday算法之类的，他们需要将该字符对应的空间存上该字符在字符串中的位置，但是数字实在太多，太浪费空间和时间用来初始化next数组，所以我们对于这种数字类型的匹配只使用KMP算法。