【题解】HDU-1711 Number Sequence

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

思路

KMP模板题

代码

查看代码
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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;
}
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