KMP(1-模板题)
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
助于理解KMP的链接:http://m.blog.csdn.net/blog/MAOTIANWANG/34466483
#include <stdio.h> #include <string.h> #include <algorithm> using namespace std; const int maxn=1000008; int t[maxn]; int s[maxn]; int n[maxn]; int len,len1; void get_next() { int i=0;//后移下标 int j=-1;//起始下标 n[0]=-1; while(i<len) { if(j==-1||t[i]==t[j]) { i++;//若匹配或为起点后移坐标后移 j++;//若匹配或为起点起始坐标后移 n[i]=j;//若匹配则记录前一起始下标情况即匹配数目,不匹配则j=0,记录0 } else j=n[j];//如果不匹配则j就倒退回之前达成匹配的最后一位,再次向后匹配,重新来过; }//当j=0时,j=n[j]=-1;便于进入if使i++,但j依旧是0,没变,让新的i,新的n[i]存储前一j的情况 } void kmp() { int i=0; int j=0; while(i<len1) { if(s[i]==t[j]||j==-1) { i++; j++; } else j=n[j]; if(j==len) { printf("%d\n",i-len+1); return ; } } printf("-1\n"); } int main() { int m; scanf("%d",&m); while(m--) { memset(n,0,sizeof(n)); scanf("%d%d",&len1,&len); for(int i=0; i<len1; i++) { scanf("%d",&s[i]); } for(int j=0; j<len; j++) { scanf("%d",&t[j]); } get_next(); kmp(); } return 0; }