# hdu 1711 Number Sequence (KMP)

xiaoxiao2021-02-28  3

# Number Sequence

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 28879    Accepted Submission(s): 12146 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 HDU 2007-Spring Programming Contest   Recommend lcy 题意：找到第一个匹配的下标。

#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> using namespace std; int t[1000005],p[10005],ne[10005]; void makeNext(const int p[],int ne[],int len) { ne[0]=0; for(int i=1,k=0;i<len;i++) { while(k>0&&p[i]!=p[k]) k=ne[k-1]; if(p[i]==p[k]) k++; ne[i]=k; } } int kmp(const int t[],const int p[],int ne[],int n,int m) { int ans=0; makeNext(p,ne,m); for(int i=0,k=0;i<n;i++) { while(k>0&&t[i]!=p[k]) k=ne[k-1]; if(t[i]==p[k]) k++; if(k==m) { ans=i-m+2; return ans; } } return -1; } int main() { int c; scanf("%d",&c); for(int i=1;i<=c;i++) { int n,m; scanf("%d%d",&n,&m); for(int i=0;i<n;i++) scanf("%d",&t[i]); for(int i=0;i<m;i++) scanf("%d",&p[i]); printf("%d\n",kmp(t,p,ne,n,m)); } return 0; }