POJ3539 Elevator 同余类BFS

Elevator

Time Limit: 4000MS $ quad $ Memory Limit: 65536K Case Time Limit: 2000MS 　 ### Description

Edward works as an engineer for Non-trivial Elevators: $ Engineering, Research and Construction (NEERC) $ .
His new task is to design a brand new elevator for a skyscraper with h floors.
　
Edward has an idée fixe: he thinks that four buttons are enough to control the movement of the elevator.
His last proposal suggests the following four buttons:
　

Move $ a $ floors up.
Move $ b $ floors up.
Move $ c $ floors up.
Return to the first floor.

Initially, the elevator is on the first floor.
A passenger uses the first three buttons to reach the floor she needs.
If a passenger tries to move $ a, b $ or $ c $ floors up
and there is no such floor (she attempts to move higher than the $ h $ -th floor), the elevator doesn’t move.
　
To prove his plan worthy, Edward wants to know how many floors are actually accessible from the first floor via his elevator.
Help him calculate this number.
　

Input

The first line of the input file contains one integer $ h $ — the height of the skyscraper $ (1 ≤ h ≤ 10^18) $ .
　
The second line contains three integers $ a, b $ and $ c $ — the parameters of the buttons $ (1 ≤ a, b, c ≤ 100,000) $ .
　

Output

Output one integer number — the number of floors that are reachable from the first floor.
　

Sample Input

 15
 4 7 9

Sample Output

Source

Northeastern Europe 2007, Northern Subregion
　

题目大意

一部电梯，最初在 $ 1 $ 层。
电梯有 $ 4 $ 个按键：上升 $ a $ 层，上升 $ b $ ，上升 $ c $，回到 $ 1 $ 层。
求从 $ 1 $ 层出发，能到达 $ 1~h $ 之间的那些楼层

题解

不妨设 $ a le b le c $ ，把楼层按照除以 $ a $ 的余数分成 $ 0,1,2,dots,a-1 $ 这 $ a $ 个类。
每个类 $ x $ 看作一个点，
向 $ (x+b) quad mod quad a $ 和 $ (x+c) quad mod quad a $ 分别连长度为 $ b,c $ 的有向边。
从 $ 1 $ 出发求单源最短路
同余类 $ x $ 中能够到达的楼层就是 $ d[x],d[x]+a,d[x]+2 imes a, dots $

代码

#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
#include<queue>
using namespace std;
#define int long long
int dis[100010],h,a,b,c,ans;
bool vis[100010];
void bfs(){
	memset(dis,0x3f,sizeof(dis)); queue<int>q;
	dis[1%a]=1; q.push(1%a); vis[1%a]=1;
	while(!q.empty()){
		int u=q.front(),v; q.pop(); vis[u]=0;
		if(dis[v=(u+b)%a]>dis[u]+b){
			dis[v]=dis[u]+b;
			if(!vis[v]) q.push(v);
		}
		if(dis[v=(u+c)%a]>dis[u]+c){
			dis[v]=dis[u]+c;
			if(!vis[v]) q.push(v);
		}
	}
}
signed main(){
	scanf("%lld %lld %lld %lld",&h,&a,&b,&c);
	if(a>b) swap(a,b); if(a>c) swap(a,c);
	bfs();
	for(int i=0;i<a;++i)
		if(dis[i]<=h) ans+=(h-dis[i])/a+1;
	printf("%lld",ans);
	return 0;
}
/*
Problem: 3539

User: potremz
Memory: 1028K

Time: 360MS
Language: C++

Result: Accepted
*/