Number Sequence

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 24863    Accepted Submission(s): 10530 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  

KMP算法的入门题目，题意为输入一个原始串和一个模板串，在原始串中找到一个连续的和模板串一样的子串，再输出其开头索引，若没有就输出-1。注意KMP停止的时候索引的位置是匹配后模版串尾端的索引，所以要减去模版串长度后+1。

#include <bits/stdc++.h> using namespace std; const int MAXN=1000010; const int MAXM=10010; int a[MAXN],b[MAXM],nex[MAXM]; int n,m; void preKMP() { int p=0,q=-1; nex[0]=-1; while(p<m){ while(q!=-1 && b[p]!=b[q]) q=nex[q]; nex[++p]=++q; } } int KMP() { preKMP(); int p=0,q=0; while(p<n&&q<m){ while(q!=-1 && a[p]!=b[q]) q=nex[q]; p++; q++; if(q==m) return p-m+1; } return -1; } int main() { int t; for(scanf("%d",&t);t;t--) { 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]); printf("%d\n",KMP()); } return 0; }