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): 15548 Accepted Submission(s): 6836
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
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1711
看了视频也看了博客,对KMP算法算是有了一知半解,先来一发水题吧。
对于next 数组 只需要求模式串,也就是两个串匹配中较小的那个。
【源代码】
#include <iostream>
#include <cstdio>
using namespace std;
const int maxn =1000005;
const int maxb =10005;
int a[maxn],b[maxb];
int nextv[maxb];
void get_next(int b[],int m){ //求模式串的 next数组
int i=0;//前缀
nextv[0]=-1;
int j=-1;//后缀
while(i<m){
if(j==-1||b[i]==b[j]){
i++; j++;
if(b[i]==b[j]) //遇到相同元素的优化
nextv[i]=nextv[j];
else
nextv[i]=j;
}
else
j=nextv[j];
}
}
int KMP(int n,int m){
int i=0;
int j=0;
while(i<n&&j<m){
if(j==-1||a[i]==b[j]){
i++;j++;
}
else{
j=nextv[j];
}
}
if(j==m){ //只有匹配出结果j才能==m
// cout<<"i: "<<i<<" "<<j<<endl;
return i-m+1;
}
else return -1;
}
int main(){
int T,n,m;
scanf("%d",&T);
while(T--){
scanf("%d%d",&n,&m);
for(int i=0;i<n;i++){
scanf("%d",&a[i]);
}
for(int i=0;i<m;i++){
scanf("%d",&b[i]);
}
get_next(b,m); //模式串 partern
int ans=KMP(n,m);
cout<<ans<<endl;
}
return 0;
}