codeforces859E Desk Disorder

最新推荐文章于 2021-11-15 22:27:18 发布

原创最新推荐文章于 2021-11-15 22:27:18 发布 · 461 阅读

2 ·

CC 4.0 BY-SA版权

文章标签：

#codeforces #并查集 #OI #计数问题 #859e

数据结构同时被 3 个专栏收录

10 篇文章

订阅专栏

暴力

9 篇文章

订阅专栏

树

5 篇文章

订阅专栏

本文介绍了一个使用并查集算法解决工程师座位重新安排的问题。面对2n个位置及每位工程师当前和期望的位置，文章详细阐述了如何通过并查集来计算所有合法座位安排的方案数，并给出具体实现代码。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

A new set of desks just arrived, and it's about time! Things were getting quite cramped in the office. You've been put in charge of creating a new seating chart for the engineers. The desks are numbered, and you sent out a survey to the engineering team asking each engineer the number of the desk they currently sit at, and the number of the desk they would like to sit at (which may be the same as their current desk). Each engineer must either remain where they sit, or move to the desired seat they indicated in the survey. No two engineers currently sit at the same desk, nor may any two engineers sit at the same desk in the new seating arrangement.

How many seating arrangements can you create that meet the specified requirements? The answer may be very large, so compute it modulo 1000000007 = 109 + 7.

Input

Input will begin with a line containing N (1 ≤ N ≤ 100000), the number of engineers.

N lines follow, each containing exactly two integers. The i-th line contains the number of the current desk of the i-th engineer and the number of the desk the i-th engineer wants to move to. Desks are numbered from 1 to 2·N. It is guaranteed that no two engineers sit at the same desk.

Output

Print the number of possible assignments, modulo 1000000007 = 109 + 7.

Examples
inputCopy
4
1 5
5 2
3 7
7 3
output
6
inputCopy
5
1 10
2 10
3 10
4 10
5 5
output
5

Note

These are the possible assignments for the first example:

1 5 3 7
1 2 3 7
5 2 3 7
1 5 7 3
1 2 7 3
5 2 7 3

题意：有2n个位置，每个人有当前的位置和想去的位置，不能冲突。求方案数，每个人可以不换。

题解：裸的并查集，我们对于每个联通块单独计算，如果是一棵树，那么我们有n种方案（每个人都可以只有他换），如果有自环，那么我们不能换，如果是个有环图，那么有两种方案（换或者不换）。

代码：

#include <stdio.h>
int f[200005],sz[200005],n,x,y,u,v,lo[200005],cyc[200005];
int find(int x){return x==f[x]?x:f[x]=find(f[x]);}
int main()
{
    scanf("%d",&n);
    for(int i=0;i<=2*n;i++) f[i]=i,sz[i]=1;
    for(int i=0;i<n;i++)
    {  
        scanf("%d%d",&x,&y);
        if(x==y) lo[find(x)]=1;
        u=find(x),v=find(y);
        if(u!=v)
        {
            f[u]=v;
            sz[v]+=sz[u];
            lo[v]|=lo[u];
        }
        else cyc[u]=1;
    }
    long long ans=1;
    for(int i=1;i<=2*n;i++)
        if(find(i)==i)
        {
            if(lo[i]) continue;
            if(cyc[i]) ans=ans*2%1000000007;
            else ans=ans*sz[i]%1000000007;
        }
    printf("%I64d\n",ans);
}