洛谷 - P3403 跳楼机(同余最短路)

Posted 2021-08-19 Frozen_Guardian

题目链接：点击查看

题目大意：求 $ax+by+cz=K,K\in [1,h]$ 有多少个 $K$ 满足方程

题目分析：设 $d[i]$ 为固定 $x$ 后能够到达 $\mathrm{m}\mathrm{o}\mathrm{d}x=i$ 的最小楼层，也就是仅通过操作 $2,3$ 所能到达的最小楼层，转移方程为 $d[i+y]=d[i]+y$ 和 $d[i+z]=d[i]+z$，此方程可以用迪杰斯特拉转移，计算出 $d$ 数组后，答案就是 $\sum (\frac{h-d[i]}{x}+1)$ 了，此处的 $+1$ 意味着原本高度为 $i$ 的楼层

代码：

// Problem: P3403 跳楼机
// Contest: Luogu
// URL: https://www.luogu.com.cn/problem/P3403
// Memory Limit: 125 MB
// Time Limit: 1000 ms
// 
// Powered by CP Editor (https://cpeditor.org)

// #pragma GCC optimize(2)
// #pragma GCC optimize("Ofast","inline","-ffast-math")
// #pragma GCC target("avx,sse2,sse3,sse4,mmx")
#include<iostream>
#include<cstdio>
#include<string>
#include<ctime>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<stack>
#include<climits>
#include<queue>
#include<map>
#include<set>
#include<sstream>
#include<cassert>
#include<bitset>
#include<list>
#include<unordered_map>
#define lowbit(x) (x&-x)
using namespace std;
typedef long long LL;
typedef unsigned long long ull;
template<typename T>
inline void read(T &x)
{
	T f=1;x=0;
	char ch=getchar();
	while(0==isdigit(ch)){if(ch=='-')f=-1;ch=getchar();}
	while(0!=isdigit(ch)) x=(x<<1)+(x<<3)+ch-'0',ch=getchar();
	x*=f;
}
template<typename T>
inline void write(T x)
{
	if(x<0){x=~(x-1);putchar('-');}
    if(x>9)write(x/10);
    putchar(x%10+'0');
}
const int inf=0x3f3f3f3f;
const int N=1e6+100;
template<typename T>
struct Dij
{
	const static int N=1e6+100;
	const static int M=1e6+100;
	struct Edge
	{
		int to,next;
		T w;
	}edge[M];
	 
	int head[N],cnt;//链式前向星 
	
	T d[N];
	 
	bool vis[N];
	 
	void addedge(int u,int v,T w)
	{
		edge[cnt].to=v;
		edge[cnt].w=w;
		edge[cnt].next=head[u];
		head[u]=cnt++;
	}
	 
	struct Node
	{
		int to;
		T w;
		Node(int TO,T W)
		{
			to=TO;
			w=W;
		}
		bool operator<(const Node& a)const
		{
			return w>a.w;
		}
	};
	 
	void Dijkstra(int st)
	{
		priority_queue<Node>q;
		memset(vis,false,sizeof(vis));
		memset(d,0x3f,sizeof(d));
		d[st]=1;
		q.push(Node(st,0));
		while(q.size())
		{
			Node cur=q.top();
			int u=cur.to;
			q.pop();
			if(vis[u])
				continue;
			vis[u]=true;
			for(int i=head[u];i!=-1;i=edge[i].next)//扫描出所有边 
			{
				int v=edge[i].to;
				T w=edge[i].w;
				if(d[v]>d[u]+w)//更新 
				{
					d[v]=d[u]+w;
					q.push(Node(v,d[v]));
				}
			}
		}
	}
	 
	void init()
	{
		memset(head,-1,sizeof(head));
		cnt=0; 
	}
};
Dij<LL>t;
int main()
{
#ifndef ONLINE_JUDGE
//	freopen("data.in.txt","r",stdin);
//	freopen("data.out.txt","w",stdout);
#endif
//	ios::sync_with_stdio(false);
	t.init();
	LL h;
	int x,y,z;
	read(h),read(x),read(y),read(z);
	if(x==1||y==1||z==1) {
		cout<<h<<endl;
		return 0;
	}
	for(int i=0;i<x;i++) {
		t.addedge(i,(i+y)%x,y);
		t.addedge(i,(i+z)%x,z);
	}
	t.Dijkstra(1);
	LL ans=0;
	for(int i=0;i<x;i++) {
		if(t.d[i]>h) {
			continue;
		}
		ans+=(h-t.d[i])/x+1;
	}
	cout<<ans<<endl;
	return 0;
}