数据大
#include<iostream>
#include<string.h>
#define maxn 1000000+5
#define maxn2 10000+5
using namespace std;
int s[maxn];
int p[maxn2];
int Next[maxn2];
void getNext(const int p[],int Next[],int plen) {
Next[0]=-1;
int k=-1;
int j=0;
while(j<plen-1) {
if(k==-1||p[k]==p[j]) {
++k;
++j;
if(p[k]!=p[j])
Next[j] = k;
else
Next[j] = Next[k];//修正nNext数组
}
else {
k = Next[k];
}
}
}
int KMP(const int s[],const int p[],int slen,int plen) {
int i=0;
int j=0;
while(i<slen&&j<plen) {
if(j==-1||s[i]==p[j]) {
i++;
j++;
} else {
j = Next[j];
}
}
if(j>=plen) {
return i-j+1;
} else {
return -1;
}
}
int main() {
int T;
scanf("%d",&T);
while(T--) {
int num1,num2;
cin>>num1;
cin>>num2;
for(int i=0; i<num1; i++)
scanf("%d",&s[i]);
for(int i=0; i<num2; i++)
scanf("%d",&p[i]);
memset(Next,-1,sizeof(Next));
getNext(p,Next,num2);
//cout<<*Next<<endl;
cout<<KMP(s,p,num1,num2)<<endl;
}
return 0;
}
http://acm.split.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): 30758 Accepted Submission(s): 12930
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