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HDU 1711 Number Sequence kmp

編輯：關於C++

Number Sequence

Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 15821 Accepted Submission(s): 6978

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

Source HDU 2007-Spring Programming Contest
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ACcode

#include

#include #include #include #include #include #include #include #define maxn 100 using namespace std; int N[1000000+10]; int M[10000+10]; int nex[10000+10]; int main(){ int loop,la,lb; scanf(%d,&loop); while(loop--){ scanf(%d%d,&la,&lb); for(int i=0;i=0&&M[pos]!=M[i-1]) pos=nex[pos]; nex[i]=++pos; } int p=0,cur=0,t=-1; while(cur=0) p=nex[p]; else { cur++; p=0; } if(p==lb){ t=cur-lb+1; break; } } printf(%d ,t); } return 0; }