Number Sequence

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 22047    Accepted Submission(s): 9420 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   |   We have carefully selected several similar problems for you:   1358  3336  1686  3746  1251    经过此题（虽是模板题）发现我这菜鸟还是对KMP算法没有深入的理解，掌握不好，处处出错。看来要继续深层次理解KMP算法啊。。。。。 #include<cstdio> #include<iostream> #include<cstring> #include<cmath> #include<algorithm> using namespace std; const int N = 1e6+10; const int M = 1e4+5; int p[N],s[N]; int ans,n,m; void getNext(int *s,int *next) { int j,k; next[0]=-1; j=0,k=-1; while(j<m-1) { if(k==-1||s[j]==s[k]) { j++; k++; // if(s[j]!=s[k]) next[j]=k; // else // next[j]=next[j]; } else k=next[k]; } } int KMPMatch(int *p,int *s) { int i,j; int next[M]; i=0,j=0; getNext(s,next); //忘加了 // printf("--\n"); while(i<n) { if(j==-1||p[i]==s[j]) { j++; i++; // printf("---\n"); } else { j=next[j]; } if(j==m) return i-m+1; } 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",&p[i]); } for(int j=0; j<m; j++) { scanf("%d",&s[j]); } ans=KMPMatch(p,s); printf("%d\n",ans); } return 0; } 题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1711