HDU 1711 Number Sequence
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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
题目大意:
给定两个数组,问能不能再第一个数组中匹配得到第二个数组,如果可以,那么输出最早匹配的起始位置,否则输出-1。
C++
#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>using namespace std;const int N=10010;int arr1[N*100],arr2[N],nt[N];int n,m,s;void getnt(){ int i = 0, j = -1; nt[0] = -1; while(i < m) { if(j == -1 || arr2[i] == arr2[j]) { i++, j++; nt[i] = j; } else j = nt[j]; }}int kmp(){ getnt(); int i = 0, j = 0; s=-1; while(i < n) { if(j == -1 || arr1[i] == arr2[j]) i++, j++; else j = nt[j]; if(j == m) { s=i-m+1; break; } } return s;}int main(){ int t; scanf("%d", &t); while(t--) { scanf("%d%d", &n, &m); for(int i = 0; i < n; i++) scanf("%d", &arr1[i]); for(int i = 0; i < m; i++) scanf("%d", &arr2[i]); if(n < m) printf("-1\n"); else cout<<kmp()<<endl; } return 0;}
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