[HDOJ1711]Number Sequence

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1711

 

 

Number Sequence

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 15683    Accepted Submission(s): 6898

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

 

 

初学KMP，水一发模版题

 

 

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <cmath>
#include <queue>
#include <map>
#include <stack>
#include <list>
#include <vector>

using namespace std;
const int maxn = 1000010;
int na, nb;
int a[maxn];
int b[maxn];
int pre[maxn];

void getpre(int *b, int *pre) {
    int j, k;
    pre[0] = -1;
    j = 0;
    k = -1;
    while(j < nb - 1) {
        if(k == -1 || b[j] == b[k]) {//匹配
            j++;
            k++;
            pre[j] = k;
        }
        else {  //b[j] != b[k]
            k = pre[k];
        }
    }
}

int kmp() {
    int i = 0;
    int j = 0;
    getpre(b, pre);
    while(i < na) {
        if(j == -1 || a[i] == b[j]) {
            i++;
            j++;
        }
        else {
            j = pre[j];
        }
        if(j == nb) {
            return i - nb + 1;
        }
    }
    return -1;
}

int main() {
    int T;
    scanf("%d", &T);
    while(T--) {
        scanf("%d %d", &na, &nb);
        for(int i = 0; i < na; i++) {
            scanf("%d", &a[i]);
        }
        for(int i = 0; i < nb; i++) {
            scanf("%d", &b[i]);
        }
        printf("%d\n", kmp());
    }
    return 0;
}

 

本题还可以用hash做，效率与kmp差距不大。

 

 

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <cmath>
#include <queue>
#include <map>
#include <set>
#include <stack>
#include <list>
#include <vector>

using namespace std;

typedef unsigned long long ull;
const int B = 100007;
const int maxn = 1000010;
int a[maxn], b[maxn];
int na, nb;

ull quickmul(int x, int n) {
    ull ans = 1;
    ull t = x;
    while(n) {
        if(n & 1) {
            ans = (ans * t);
        }
        t = t * t;
        n >>= 1;
    }
    return ans;
}

int contain() {
    if(na > nb) {
        return false;
    }
    ull t = quickmul(B, na);
    ull ah = 0, bh = 0;
    for(int i = 0; i < na; i++) {
        ah = ah * B + a[i];
    }
    for(int i = 0; i < na; i++) {
        bh = bh * B + b[i];
    }
    for(int i = 0; i + na <= nb; i++) {
        if(ah == bh) {
            return i + 1;
        }
        if(i + na < nb) {
            bh = bh * B + b[i+na] - b[i] * t;
        }
    }
    return -1;
}
int main() {
    // freopen("in", "r", stdin);
    int T;
    scanf("%d", &T);
    while(T--) {
        scanf("%d %d", &nb, &na);
        for(int i = 0; i < nb; i++) {
            scanf("%d", &b[i]);
        }
        for(int i = 0; i < na; i++) {
            scanf("%d", &a[i]);
        }
        printf("%d\n", contain());
    }
    return 0;
}