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

從頭到尾徹底理解KMP：
https://www.cnblogs.com/zhangtianq/p/5839909.html

#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;
}