HDU 1711-Number Sequence

Number Sequence

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 31779    Accepted Submission(s): 13350


题目链接：点击打开链接

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

题意：

给你两个数组 a[1], a[2], ...... , a
, 和b[1], b[2], ...... , b[M] ，N>M,让你找到一个K,使得 a[K] = b[1], a[K + 1] = b[2], ......
, a[K + M - 1] = b[M].说白了，，就是让你找到第二个串在第一个串有没有与之匹配的，若有，则返回 a 数组中匹配的初始位置，否则输出-1.

分析：

KMP算法，今天花了两个小时自己模拟了一遍，终于完全明白了，等整理好今天下午写的几题之后，就来写写KMP的详解，一个模板，稍加修改，写了四个题。（这里分享一个要注意的地方，就是如果你用的数组名字叫next的话，就不要出现c++的头文件，提交会错，因为和c++里的东西重名了）

#include<stdio.h>
#include<string>
#include<string.h>
const int INF1=10005;
const int INF2=1000005;
int a[INF2],b[INF1];
int next[INF1];
int m,n;

void get_next()/// 求出模式串的next数组
{
next[0] = -1;
int k = -1;
for (int i = 1; i < m; i++)
{
while (k > -1 && b[k + 1] != b[i])
{
k = next[k];
}
if (b[k + 1] == b[i])
{
k = k + 1;
}
next[i] = k;
}
}

int KMP()
{
get_next();
int k = -1;
for (int i = 0; i < n; i++)
{
while (k >-1&& b[k + 1] != a[i])
{
k = next[k];
}
if (b[k + 1] == a[i])
k = k + 1;
if (k == m-1)
{
return (i-m+2);///因为我的下标是从0开始的，所以应该加2
}
}
return -1;
}

int main()
{
int t;
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]);
int k=KMP();
printf("%d\n",k);
}
return 0;
}