字符串_KMP算法(求next[]模板 hdu 1711)

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

问题描述：给两个序列a,b，长度分别为n,m(1<=n<=1000000,1<=m<=10000),问序列b是否为序列a的子序列，若是：返回a中最左边的与b相等的子序列的首元素下标；若不是，输出-1。

目的：方便以后查看KMP算法中next[]的模板

Number Sequence

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 12811    Accepted Submission(s): 5815

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

代码实现：

#include "stdio.h"
#include "string.h"
#define N 10005

int next[N];
int a[100*N],b[N];

void KMP(int *s,int len,int *next) //求next[]模板
{
    int i,j;
    i = 0;
    j = next[0] = -1;
    while(i<len)
    {
        while(j!=-1 && s[i]!=s[j])
            j = next[j];
        next[++i] = ++j;
    }
}

int main()
{
    int T;
    int i,j;
    int n,m;
    scanf("%d",&T);
    while(T--)
    {
        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]);
        KMP(b,m,next);
        i=j=0;
        while(i<n)
        {
            while(j!=-1 && a[i]!=b[j])
                j = next[j];
            i++,j++;
            if(j==m) break;
        }
        if(j==m) printf("%d
",i-j+1); //题中数据是从下标为1开始的，故加1
        else printf("-1
");
    }
    return 0;
}