HDU-1711 Number Sequence kmp算法

题目链接

http://acm.hdu.edu.cn/showproblem.php?pid=1711

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题目

Number Sequence
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 32182 Accepted Submission(s): 13497

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

题意

给定两个整数数组，判断b数组是否在a数组中出现，并计算出现的位置。

题解

将整数看作字符，那么就变成了一个求字符串匹配问题了。分析n最大有1e6那么大，所以暴力匹配肯定不行啦。
这里采用kmp算法，时间复杂度O(n+m).
没学过kmp，戳这里http://blog.csdn.net/qq_37685156/article/details/78816196

代码

#include <iostream>#include <cstdio>#include <cstring>using namespace std;const int maxn=1e6+100;const int maxm=1e4+100;int a[maxn],b[maxm],n,m,pre[maxm];void getpre(){    int j=0;    memset(pre,0,sizeof(pre));    for(int i=2;i<=m;i++)    {        while(j>0 && b[j+1]!=b[i]) j=pre[j];        if(b[j+1]==b[i]) j++;        pre[i]=j;    }}int kmp(){    int j=0;    for(int i=1;i<=n;i++)    {        while(j>0 && b[j+1]!=a[i]) j=pre[j];        if(b[j+1]==a[i]) 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=1; i<=n; i++)            scanf("%d",&a[i]);        for(int i=1; i<=m; i++)            scanf("%d",&b[i]);        getpre();        printf("%d\n",kmp());    }    return 0;}