HDU 1711 Number Sequence
http://acm.hdu.edu.cn/showproblem.php?pid=1711
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
代码:
/* pku3461(Oulipo), hdu1711(Number Sequence) 这个模板 字符串是从0开始的 Next数组是从1开始的 */ #include <iostream> #include <stdio.h> #include <string.h> #include <algorithm> #include <math.h> using namespace std; const int maxn = 2e7 + 5; int N, M; int nx[maxn]; int a[maxn], b[maxn]; void getNext() { int j, k; j = 0; k = -1; nx[0] = -1; while(j < M) if(k == -1 || b[j] == b[k]) nx[++j] = ++k; else k = nx[k]; } /* 返回模式串T在主串S中首次出现的位置 返回的位置是从0开始的。 */ int KMP_Index() { int i = 0, j = 0; getNext(); while(i < N && j < M) { if(j == -1 || a[i] == b[j]) { i++; j++; } else j = nx[j]; } if(j == M) return i - M + 1; else return -1; } int main() { int TT; cin>>TT; while(TT--) { 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]); int ans = KMP_Index(); printf("%d\n", ans); } return 0; } /* test case aaaaaa a abcd d aabaa b */
对着 kuangbin 的 KMP 模板写下来的题目!板子里的 next 数组改成了 nx 一直报 CE 让我一度怀疑是不是智商不行 按着板子写都会写错 QAQ