HDU 1711 Number Sequence (KMP 入门)
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Number Sequence
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
题目大意:给定两个序列a,b,求出b序列在a序列中第一次出现的位置,若不在a序列中输出-1;
思路:KMP模板直接套用,这里觉得这位博主写的挺通俗易懂的还是很简单的就看懂了 https://blog.csdn.net/starstar1992/article/details/54913261/
1 #include<iostream> 2 #include<algorithm> 3 #include<cstring> 4 #include<string> 5 6 using namespace std; 7 const int maxn = 1000005; 8 int rs[4][4], nex[maxn], a[maxn], b[maxn/100 + 5], n, m; 9 void cal_next(int *str, int len) 10 { 11 nex[0] = -1; int k = -1; 12 for (int i = 1; i <= len - 1; i++) { 13 while (k > -1 && str[k + 1] != str[i]) 14 k = nex[k]; 15 if (str[k + 1] == str[i])k = k + 1; 16 nex[i] = k; 17 } 18 } 19 int KMP(int *a, int la, int *b, int lb) 20 { 21 cal_next(b , lb); 22 int k = -1; 23 for (int i = 0; i < la; i++) { 24 while (k > -1 && b[k + 1] != a[i]) 25 k = nex[k]; 26 if (b[k + 1] == a[i])k = k + 1; 27 if (k == lb - 1)return i - lb + 1; 28 } 29 return -1; 30 } 31 int main() 32 { 33 ios::sync_with_stdio(false); 34 int T; 35 cin >> T; 36 while (T--) { 37 cin >> n >> m; 38 for (int i = 0; i < n; i++)cin >> a[i]; 39 for (int i = 0; i < m; i++)cin >> b[i]; 40 int ans = KMP(a, n, b, m); 41 if (ans != -1)cout << ans + 1 << endl; 42 else cout << "-1" << endl; 43 } 44 return 0; 45 }
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