题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1711
Number Sequence
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 54826 Accepted Submission(s): 21988
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
Recommend
lcy
题目意思:就是首先输入t表示要处理数据的组数,然后输入n和m,后面跟着两个长度为n和m的数组。找到长度为m的数组在长度为n数组里面的出现位置(位置要最小)。
题目思路:只需要在输入的地方稍微处理一下(原先的字符串输入改为数字输入)就行,然后直接kmp。
ac代码:
#include<stdio.h> #include<string> const int maxn=1e6+10; const int maxm=1e4+10; int t,n,m,s[maxn],p[maxn]; int next[maxm]; void GetNext() { int plen=0; int slen=-1; next[0]=-1; while(plen<m) { if(slen==-1||p[plen]==p[slen]) { plen++;slen++; if(p[plen]!=p[slen])next[plen]=slen; else next[plen]=next[slen]; } else slen=next[slen]; } } int kmp() { int plen=0; int slen=0; while(plen<m&&slen<n) { if(plen==-1||p[plen]==s[slen]) { plen++;slen++; } else plen=next[plen]; } if(plen==m) return slen-plen+1; else return -1; } int main() { scanf("%d",&t); while(t--) { scanf("%d%d",&n,&m); for(int i=0;i<n;i++)scanf("%d",&s[i]); for(int i=0;i<m;i++)scanf("%d",&p[i]); GetNext(); printf("%d ",kmp()); } return 0; }