Number Sequence
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 40427 Accepted Submission(s): 16672
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
Source
HDU 2007-Spring Programming Contest
#include <bits/stdc++.h>
using namespace std;
const int mn = 1000010, mm = 10010;
int n, m;
int a[mn], b[mm];
int nx[mm];
void cal_next(int b[])
{
nx[0] = -1;
int k = -1;
for (int i = 1; i <= m - 1; i++)
{
while (k > -1 && b[k + 1] != b[i])
k = nx[k];
if (b[k + 1] == b[i])
k++;
nx[i] = k;
}
}
int KMP(int a[], int b[]) /// 在 a 中查找 b
{
cal_next(b);
int k = -1;
for (int i = 0; i < n; i++)
{
while (k > -1 && b[k + 1] != a[i])
k = nx[k];
if (b[k + 1] == a[i])
k++;
if (k == m - 1)
{
return i - m + 2;
// 可重叠继续寻找下一个则不return
// 不可重叠则 k = -1;
}
}
return -1;
}
int main()
{
int T;
scanf("%d", &T);
while (T--)
{
scanf("%d %d", &n, &m);
for (int i = 0; i < n; i++) // 数组从0开始
scanf("%d", &a[i]);
for (int i = 0; i < m; i++)
scanf("%d", &b[i]);
printf("%d\n", KMP(a, b));
}
return 0;
}