HDU 1711 Number Sequence
2014-04-30 18:39
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Number Sequence
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 9899 Accepted Submission(s): 4518
[align=left]Problem Description[/align]
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.
[align=left]Input[/align]
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].
[align=left]Output[/align]
For each test case, you should output one line which only contain K described above. If no such K exists, output -1 instead.
[align=left]Sample Input[/align]
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
[align=left]Sample Output[/align]
6
-1
Kmp:
[code]#include<cstdio> #include<string> #include<cstring> #include<iostream> #include<algorithm> #define MAXN 1000010 int T[MAXN], P[MAXN/10], fail[MAXN/10], N, M; void getFail(){ fail[0] = -1; for(int i = 1, j = -1;i < M;i ++){ while(P[i] != P[j+1] && j >= 0) j = fail[j]; if(P[i] == P[j+1]) j++; fail[i] = j; } } void Kmp(){ getFail(); for(int i = 0, j = 0;i < N;i ++){ while(T[i] != P[j] && j) j = fail[j-1] + 1; if(T[i] == P[j]) j++; if(j == M){ printf("%d\n", i-M+2); return; } } printf("-1\n"); } int main(){ int t; /* freopen("in.c", "r", stdin); */ scanf("%d", &t); while(t--){ scanf("%d%d", &N, &M); for(int i = 0;i < N;i ++) scanf("%d", T+i); for(int i = 0;i < M;i ++) scanf("%d", P+i); Kmp(); } return 0; }
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