Code Lock（并查集 + 快速幂）

最新推荐文章于 2022-07-23 16:19:40 发布

原创最新推荐文章于 2022-07-23 16:19:40 发布 · 418 阅读

1 ·

CC 4.0 BY-SA版权

文章标签：

#并查集

并查集专栏收录该内容

7 篇文章

订阅专栏

本文介绍了一个名为CodeLock的问题，该问题涉及一系列轮盘组成的锁，每个轮盘上有26个英文字母。通过特定的连续轮盘操作改变锁的状态，讨论了如何计算不同锁状态的数量，并提供了一种算法解决方案。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Code Lock
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 277 Accepted Submission(s): 127

Problem Description
A lock you use has a code system to be opened instead of a key. The lock contains a sequence of wheels. Each wheel has the 26 letters of the English alphabet ‘a’ through ‘z’, in order. If you move a wheel up, the letter it shows changes to the next letter in the English alphabet (if it was showing the last letter ‘z’, then it changes to ‘a’).
At each operation, you are only allowed to move some specific subsequence of contiguous wheels up. This has the same effect of moving each of the wheels up within the subsequence.
If a lock can change to another after a sequence of operations, we regard them as same lock. Find out how many different locks exist?

Input
There are several test cases in the input.

Each test case begin with two integers N (1<=N<=10000000) and M (0<=M<=1000) indicating the length of the code system and the number of legal operations.
Then M lines follows. Each line contains two integer L and R (1<=L<=R<=N), means an interval [L, R], each time you can choose one interval, move all of the wheels in this interval up.

The input terminates by end of file marker.

Output
For each test case, output the answer mod 1000000007

Sample Input
1 1
1 1
2 1
1 2

Sample Output
1
26

解题思路
有n个字母执行m个操作然后问如果考虑这n个字母则有26的n次方种然而题目要求中又说了当执行之种可执行的操作则认为是统一序列所以只需要用N - 可执行的序列种数的26 次方即可
而这里需要注意的是[1,3][3,5][1,5]这是三把不同的锁因为3执行了两次
而[1,3][4,5][1,5] 则是两把锁
所以我们在合并时只需要合并[l-1,r] 这个区间即可
然后直接用一个快速幂算法求出来即可

#include<iostream>
#include<algorithm>
#include<string>
#include<string.h>
#include<stdio.h>

using namespace std;

long long mod = 1000000007;
long  ff[10000007];
long num =0;
long father(int x)
{
     int i = x;
    while(x!=ff[x])
        x = ff[x];

   /* while(i!=x)//压缩路径
    {
        int j = ff[i];
        ff[i] = x;
        i = j;
    }*/
    return x;
}
void Union_set(int l,int r)
{
    long pa = father(l);
    long pb = father(r);
    if(pa!=pb)
    {
        ff[pa] = pb;
        num++;
    }
}
long long exp(long n)//快速幂算法
{
    long long sum =1;
    long long tmp =26;
    while(n)
    {
        if(n%2)//
        {
            sum = sum * tmp;
            sum %= mod;
        }
        tmp = (tmp *tmp) % mod;
        n = n /2;//向右移动一位
    }
    return sum;
}
int main()
{
    int n,m;
    while(cin>>n>>m)
    {
        num =0;
        for(int i =0;i<=n;i++)//初始化 这列由于我判断是l-1 所以最左边是0
            ff[i] = i;
        for(int i =0;i<m;i++)
        {
            int l,r;
            cin>>l>>r;
            Union_set(l-1,r);//这里只需要判断Union_set(l-1,r) 或者 Union_set(l,r+1) 即可
        }
        cout<<exp(n-num)%mod<<endl;
    }
    return 0;
}



/*
1 1
1 1
2 1
1 2
*/