HDU 1711 Number Sequence (数字KMP)

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Number Sequence

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 12256    Accepted Submission(s): 5593

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
213 51 2 1 2 3 1 2 3 1 3 2 1 21 2 3 1 313 51 2 1 2 3 1 2 3 1 3 2 1 21 2 3 2 1
 

Sample Output
6-1
 

Source
HDU 2007-Spring Programming Contest

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

题目大意:给两个数组,求如果第二个数组序列在第一个数组中出现了,求在第一个数组中第一次出现的位置

题目分析:直接对数字kmp就行了,其实就和字符串kmp一样,算法的原理是一样的,但是实在不明白为什么改进的失败函数费时必未改进的还多

#include <cstdio>#include <cstring>int const MAX1 = 1e6 + 5;int const MAX2 = 1e4 + 5;int s1[MAX1], s2[MAX2];int next[MAX2], n, m;void get_next(){int i = 0, j = -1;next[0] = -1;while(i < m){if(j == -1 || s2[i] == s2[j]){i++;j++;// if(s2[i] == s2[j])// next[i] = next[j];// elsenext[i] = j;}elsej = next[j];}}int KMP(){get_next();int i = 0, j = 0;while(i < n){if(j == -1 || s1[i] == s2[j]){i++;j++;}elsej = next[j];if(j == m)return i - j + 1;}return -1;}int main(){int T;scanf("%d", &T);while(T--){int tmp;scanf("%d %d", &n, &m);if(n < m){printf("-1\n");continue;}for(int i = 0; i < n; i++)scanf("%d", &s1[i]);for(int i = 0; i < m; i++)scanf("%d", &s2[i]);int ans = KMP();printf("%d\n", ans);}}


 

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