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
给出两个数字组成的序列a,b，求b在a的哪个位置开始能完全匹配，有多个输出最小的 对KMP算法稍微一改，如果匹配串能够遍历完，则直接返回当前原串的位置，而不是记录下个数
#include <iostream> #include <cstring> #include <string> #include <vector> #include <queue> #include <cstdio> #include <set> #include <math.h> #include <algorithm> #include <queue> #include <iomanip> #include <ctime> #define INF 0x3f3f3f3f #define MAXN 10005 #define Mod 1000000007 using namespace std; int a[MAXN*100],b[MAXN]; int nextb[MAXN],n,m; void getnext() { int i=0,j=-1; nextb[0]=-1; while(i<m) { if(j==-1||b[i]==b[j]) { i++; j++; nextb[i]=j; } else j=nextb[j]; } } int kmp() { int i=0,j=0; while(i<n) { if(j==-1||a[i]==b[j]) { i++; j++; } else j=nextb[j]; if(j==m) return i; } return -1; } int main() { int t; scanf("%d",&t); while(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]); getnext(); int ans=kmp(); if(ans==-1) printf("%d\n",ans); else printf("%d\n",ans-m+1); } return 0; }