HDU 1711 KMP算法模板

Number Sequence
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 15606 Accepted Submission(s): 6868

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

练习kmp算法

#include <iostream>#include <algorithm>#include <cstdio>#include <cstring>//#define LOCALusing namespace std;int a[1000010];int b[10010];int c[10010];void getFail(int len);int Find(int n,int m);int main(){#ifdef LOCAL    freopen("data.in","r",stdin);#endif // LOCAL    int N;    scanf("%d",&N);    while(N--){        int m,n;        int i,j;        scanf("%d%d",&n,&m);        for(i = 0;i<n;i++)            scanf("%d",&a[i]);        for(i = 0;i<m;i++)            scanf("%d",&b[i]);        getFail(m);        printf("%d\n",Find(n,m));    }    return 0;}int Find(int n,int m){    int i,j = 0;    for(i = 0;i<n;i++){        while(j&&a[i]!=b[j])            j = c[j];        if(a[i] == b[j])            j++;        if(j == m)            return i+2-m;    }    return -1;}void getFail(int len){    c[0] = c[1] = 0;    int i,j;    for(i = 1;i<len;i++){        j = c[i];        while(j&&b[i]!=b[j])            j = c[j];        c[i+1] = b[i] == b[j]?j+1:0;    }}