題目

Number Sequence

Time Limit: 10000/5000 MS (Java/Others)????Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 40385????Accepted Submission(s): 16656



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

代碼

#include <iostream> #include <algorithm> #include <cstdio> #include <cstring> using namespace std; const int MAXN=1e6 + 10; int a[MAXN],b[MAXN]; int nxt[MAXN]; int n,m; void getnxt() {int i=1;int j=0;nxt[0]=0;while(i<m){if(b[i]==b[j]){nxt[i++]=++j;}else if(!j){i++;}else{j=nxt[j-1];}} } int kmp() {int i=0;int j=0;while(i<n && j<m){if(a[i]==b[j]){i++;j++;}else if(!j){i++;}else{j=nxt[j-1];}}if(j==m) return i-m+1;else return -1; } int main() {int t,i,k; scanf("%d",&t);while(t--){scanf("%d%d",&n,&m);for(i=0;i<n;i++)scanf("%d",&a[i]);for(i=0;i<m;i++)scanf("%d",&b[i]);getnxt();cout<<kmp()<<endl;}return 0; }

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