HDU - 1711 Number Sequence

最新推荐文章于 2021-07-20 09:18:01 发布

Lngxling

最新推荐文章于 2021-07-20 09:18:01 发布

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分类专栏： kmp

本文链接：https://blog.youkuaiyun.com/Lngxling/article/details/78073047

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本文解析了一道经典的KMP算法模板题，题目要求在一大序列中查找一小序列出现的位置，采用KMP算法进行高效匹配。文章提供了完整的C++代码实现，并通过样例输入输出验证了算法的有效性。

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Number Sequence

Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 30569 Accepted Submission(s): 12863

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

kmp的模板题可以看我转载的一篇讲kmp的博客讲的还是蛮好的 http://blog.youkuaiyun.com/lngxling/article/details/78072995

用int储存可以过用char就WA 不知道为啥

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <queue>
#include <stack>
#include <map>
#include <cmath>
#include <vector>
#define max_ 1000100
#define inf 0x3f3f3f3f
#define ll long long
using namespace std;
int a[max_],b[10010];
int n,m;
int f[max_];
int main(int argc, char const *argv[])
{
	int t;
	scanf("%d",&t);
	while(t--)
	{
		scanf("%d%d",&n,&m);
		int i,j,k;
		for(i=0;i<n;i++)
			scanf("%d",&a[i]);
		for(i=0;i<m;i++)
			scanf("%d",&b[i]);
		f[0]=-1;
		for(i=1;i<m;i++)
		{
			j=f[i-1];
			while(a[j+1]!=b[i]&&j>=0)
				j=f[j];
			if(a[j+1]==b[i])
				f[i]=j+1;
			else
				f[i]=-1;
		}
		i=0,j=0;
		int flag=0;
		while(i<n)
		{
			if(a[i]==b[j])
			{
				i++;
				j++;
				if(j==m)
				{
					printf("%d\n",i-m+1);
					flag=1;
					break;
				}

			}
			else
			{
				if(j==0)
					i++;
				else
					j=f[j-1]+1;
			}
		}
		if(flag==0)
			printf("-1\n");
	}
	return 0;
}