HDOJ 1711Number Sequence

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 15622    Accepted Submission(s): 6877


Problem Description

Given two sequences of numbers : a[1], a[2], ...... , a
, 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
.
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 <cstdio>

const int N = 1000010;
const int M = 10010;
int ary
;
int p[M], next[M];
int n, m;

void get_next() {
int i = 0, j = -1;
next[0] = -1;
while (i < m) {
if (j == -1 || p[i] == p[j]) {
++i, ++j;
if (p[i] != p[j])
next[i] = j;
else
next[i] = next[j];
}
else
j = next[j];
}
}

int KMP() {
int i = 0, j = 0;
while (i < n && j < m) {
if (ary[i] == p[j])
++i, ++j;
else {
if (next[j] == -1)
++i, j = 0;
else
j = next[j];
}
}
if (j == m)
return i - j + 1;
else
return -1;
}

int main() {
int T;
scanf("%d", &T);

while (T--) {
scanf("%d%d", &n, &m);
for (int i = 0; i < n; ++i)
scanf("%d", ary + i);
for (int i = 0; i < m; ++i)
scanf("%d", p + i);
get_next();
printf("%d\n", KMP());
}

return 0;
}