hdu1711 Number Sequence kmp应用
2016-08-02 22:42
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题目地址:http://acm.hdu.edu.cn/showproblem.php?pid=1711
题目:
Problem Description 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.
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
. 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
思路:裸的kmp应用,直接用kmp就好了,就当练手
代码:
#include <bits/stdc++.h> #define PB push_back #define MP make_pair using namespace std; typedef long long LL; typedef pair<int,int> PII; #define PI acos((double)-1) #define E exp(double(1)) #define K 1000000+9 int n,m; int nt[K],a[K],b[K]; void kmp_next(void) { memset(nt,0,sizeof(nt)); for(int i=1,j=0;i<m;i++) { while(j&&b[i]!=b[j])j=nt[j-1]; if(b[i]==b[j])j++; nt[i]=j; } } int kmp(void) { if(m>n)return -1; kmp_next(); for(int i=0,j=0;i<n;i++) { while(j&&a[i]!=b[j])j=nt[j-1]; if(a[i]==b[j])j++; if(j==m) return i-m+2; } return -1; } int main(void) { int t;cin>>t; while(t--) { cin>>n>>m; for(int i=0;i<n;i++) scanf("%d",&a[i]); for(int i=0;i<m;i++) scanf("%d",&b[i]); printf("%d\n",kmp()); } return 0; }
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