HDU 1711 Number Sequence ——KMP

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

##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

KMP的使用

从头到尾彻底理解KMP:
https://www.cnblogs.com/zhangtianq/p/5839909.html

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#include <iostream>
#include <cstdio>
#include <cstring>

//AC

//KMP
using namespace std;


int q[1000005];
int sub[10005];
int nextsub[10005];
int t, n, m;


int main(int argc, char const *argv[])
{
scanf("%d", &t);

while (t--) {

scanf("%d%d", &n, &m);

for (int i = 0; i < n; i++) {
scanf("%d", q + i);
}
for (int i = 0; i < m; i++) {
scanf("%d", sub + i);
}

nextsub[1] = 0;

for (int i = 1; i < m; i++) {
int cpi = i;
while (1) {
if (cpi == 0) {
nextsub[i + 1] = 0;
break;
}

if (sub[nextsub[cpi]] == sub[cpi]) {
nextsub[i + 1] = nextsub[cpi] + 1;
break;
} else {
cpi = nextsub[cpi];
}
}
}

int i = 0, j = 0, find = 0;
while (i < n) {
if (q[i] == sub[j]) {
i++;
j++;

if (j == m) {
find = 1;
printf("%d\n", i - m + 1);
break;
}
} else {
if (j == 0) {
i++;
}

j = nextsub[j];
}
}
if (!find) {
printf("-1\n");
}
}

return 0;
}