杭电 1711 KMP算法

是一道KMP的水题，，，裸题，，，就是一个简单的模板应用。。。。。。题目：

Number Sequence

Time Limit: 10000/5000 MS (Java/Others)Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3689Accepted Submission(s): 1688

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

ac代码：用的G++，跑了1000多ms，，，求优化。

#include <iostream> #include <string> #include <cstdio> #include <string.h> using namespace std; const int N=1000005,M=100005; int a[N],b[M],nextt[M]; int n,m; void getnext(){ int i=1,j=0; nextt[1]=0; while(i<m){ if(j==0||b[i]==b[j]){ ++i;++j; if(b[i]!=b[j]) nextt[i]=j; else nextt[i]=nextt[j]; } else j=nextt[j]; } } int kmp(){ int i=1,j=1; while(i<=n&&j<=m){ if(j==0||a[i]==b[j]){ ++i;++j; } else j=nextt[j]; } if(j>m) return i-m; return 0; } int main(){ int kk; scanf("%d",&kk); while(kk--){ memset(a,0,sizeof(a)); memset(b,0,sizeof(b)); memset(nextt,0,sizeof(nextt)); //int n,m; scanf("%d%d",&n,&m); for(int i=1;i<=n;++i) scanf("%d",&a[i]); for(int i=1;i<=m;++i) scanf("%d",&b[i]); getnext(); int flag=kmp(); if(flag==0) printf("-1\n"); else printf("%d\n",flag); } return 0; }