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  • POJ3539 Elevator 同余类BFS

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    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

     9
    

     

    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
    */
    
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  • 原文地址:https://www.cnblogs.com/PotremZ/p/POJ3539.html
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