HDU 1711 Number Sequence(KMP)
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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
思路:很明显的kmp题
#include<bits/stdc++.h> using namespace std; #define ll long long const ll inf = 2000000000000; const int mod = 1000000007; const int maxn = 1000 + 8; int n, m, k, t, ne[10000 + 8]; int a[1000000 + 8], b[10000 + 8]; void getnext() { ne[0] = ne[1] = 0; for(int i = 1; i < m; i++) { int j = ne[i];/// while(j && b[i] != b[j]) j = ne[j]; ne[i + 1] = b[i] == b[j] ? j + 1 : 0;///如果两个字母相同,则i和j都加加;否则j = 0,重新比较 } } int kmp() { int j = 0; for(int i = 0; i < n; i++) { while(j && b[j] != a[i])///如果匹配不成功 j = ne[j]; if(a[i] == b[j]) j++; if(j == m) return i - m + 2; } return -1; } int main() { scanf("%d", &t); while(t--) { memset(ne, 0, sizeof(ne)); scanf("%d%d", &n, &m); for(int i = 0; i < n; i++) scanf("%d", &a[i]); for(int i = 0; i < m; i++) scanf("%d", &b[i]); getnext(); k = kmp(); printf("%d ", k); } return 0; }
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