hdu_1711Number Sequence(kmp)
2016-07-16 15:03
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Number Sequence
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 20445 Accepted Submission(s): 8756
[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
赤裸裸的kmp
//kmp #include<cstdio> #include<cstring> #include<algorithm> using namespace std; int a[1000005],b[10005]; int Next[10005]; int n,m; int kmp() { int i,j; j = 0; int tm = Next[0] = -1; //ÇóNextÊý×é while(j<m-1){ if(tm<0||b[j]==b[tm]) Next[++j] = ++tm; else tm = Next[tm]; } //Æ¥Åä for( i = j = 0; i < n&&j < m; ){ if(j<0||a[i]==b[j])i++,j++; else j = Next[j]; } if(j<m) return -1; return i-j; } int main() { int T; scanf("%d",&T); while(T--) { scanf("%d%d",&n,&m); for(int i = 0; i< n; i++) scanf("%d",&a[i]); for(int i = 0; i < m; i++) scanf("%d",&b[i]); int ans = kmp(); if(ans!=-1) printf("%d\n",ans+1); else puts("-1"); } return 0; }
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