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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[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
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;
}
hdu1711 Number Sequence kmp应用
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原文地址：http://www.cnblogs.com/weeping/p/5731036.html
