HDU 1711 Number Sequence（KMP模板）

Number Sequence

题目

Time Limit:5000MS Memory Limit:32768KB

• 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

分析

标准到不能更标准的KMP模板（其实这里是MP算法，KMP还需要对失配函数进行优化）。

代码

#include 
#include 
#include 
#include 
using namespace std;
int a[1000005];
int b[10005];
int f[10005];
int n, m;
void getFail()
{
    f[0] = 0; f[1] = 0;
    for(int i=1; iint j =f[i];
        while(j && b[i]!=b[j])
            j = f[j];
        f[i+1] = b[i]==b[j] ? j+1 : 0;
    }
}
int kmp()
{
    getFail();
    int j = 0;
    for(int i=0; iwhile(j && b[j]!=a[i])
            j =f[j];
        if(b[j]==a[i])
            j++;
        if(j == m)
            return i-m+2;
    }
    return -1;
}
int main()
{
    int T;
    scanf("%d", &T);
    while(T--)
    {
        scanf("%d%d", &n, &m);
        for(int i=0; iscanf("%d", a+i);
        for(int i=0; iscanf("%d", b+i);
        printf("%d\n",kmp());
    }
    return 0;
}