【题解】HDU-1711 Number Sequence

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HDU-1711 Number Sequence 类型 字符串 KMP

Number Sequence (HDU-1711)

题面

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.

输入

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].

输出

For each test case, you should output one line which only contain K described above. If no such K exists, output -1 instead.

样例输入

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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

样例输出

1
2
6
-1

提示

无

思路

KMP模板题

代码

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int s[mxn], t[mxn];
int nxt[mxn];

void getnxt(int* t, int m)
{
    int i = 0, j = -1; nxt[0] = -1;
    while (i < m)
    {
        if (j == -1 || t[i] == t[j]) {
            i++, j++;
            // if (t[i] == t[j])
            //     nxt[i] = nxt[j]; // next数组优化
            // else
                nxt[i] = j;
        } else
            j = nxt[j];
    }
}

int KMP(int* s, int* t, int n, int m)
{
    int i = 0, j = 0, ans = 0;
    while (i < n)
    {
        if (j == -1 || s[i] == t[j]) {
            i++, j++;
            if (j >= m) {   // 匹配
                // ans++;
                // j = nxt[j];
                return i-j;
            }
        } else
            j = nxt[j];
    }
    // return ans;
    return -1;
}

int main()
{
    int T; scanf("%d", &T);
    while(T--)
    {
        int n, m;
        scanf("%d %d", &n, &m);
        for(int i=0; i<n; i++) scanf("%d", &s[i]);
        for(int i=0; i<m; i++) scanf("%d", &t[i]);
        getnxt(t, m);
        int ans = KMP(s, t, n, m);
        printf("%d\n", ans == -1 ? -1 : ans+1);
    }
    return 0;
}