HDU 1711 Number Sequence(KMP入门)

Number Sequence

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 6786    Accepted Submission(s): 3084

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

213 51 2 1 2 3 1 2 3 1 3 2 1 21 2 3 1 313 51 2 1 2 3 1 2 3 1 3 2 1 21 2 3 2 1

 

Sample Output

6-1

 

Source

HDU 2007-Spring Programming Contest

 

  hdu kmp入门题，一开始看到这个题目，看了数据范围，就直接否定了自己最开始想到的算法O(m*n)。想着用stl string中的find，网上查了一下string中find时间复杂度是O(m*n)的，接下来就是补基础，看《算法导论》，一知半解的，根据书上的伪代码勉强AC了，跑了500ms，膜拜一下跑进100ms的牛牛，希望能得到各路大虾们的指点，俺也优化优化。

#include <stdio.h>#define nMax 1000002int a[nMax],b[nMax];//a为文本，b为模式串 int s[nMax]; //前缀函数 int n,m;void compute(){//计算前缀函数 int k=0,q;s[1]=0;for(q=2;q<=m;q++){while(k>0 && b[k+1]!=b[q]){k=s[k];}if(b[k+1]==b[q])k=k+1;s[q]=k;}}void kmpmatch(){int q=0,i,j;//q记录已经匹配到的模式串中的个数 for(i=1;i<=n;i++){//从左至右逐个匹配文本中的字符 while(q>0&&b[q+1]!=a[i]){q=s[q];//下一个字符未能匹配，通过前缀函数计算下个文本匹配的位置}if(b[q+1]==a[i])//下一个字符能匹配 q=q+1;if(q==m){//模式串中字符全部匹配 printf("%d\n",i-m+1);return;}}printf("-1\n");return;}int main(int argc, char *argv[]){int cas,i,j;scanf("%d",&cas);while(cas--){scanf("%d%d",&n,&m);for(i=1; i<=n;i++){ scanf("%d",&a[i]); } for(i=1; i<=m;i++){ scanf("%d",&b[i]); } compute(); kmpmatch();} return 0;}