Znzryb's Blog
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Posted on Edited on

思路讲解

多源广搜经典套路，如果该点到达时间没有别的点早，就可以不用走了

然后把所有点加入q，实现多源广搜

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FOR(i,1,N){
	FOR(j,1,M){
		if(maze[i][j]!='H') continue;
		q.push({i,j,0});
		vis[i][j]=0;
	}
}
bfs();

AC代码

https://atcoder.jp/contests/abc383/submissions/62366428

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// Problem: C - Humidifier 3
// Contest: AtCoder - Daiwa Securities Co. Ltd. Programming Contest 2024（AtCoder Beginner Contest 383）
// URL: https://atcoder.jp/contests/abc383/tasks/abc383_c
// Memory Limit: 1024 MB
// Time Limit: 2000 ms
// 
// Powered by CP Editor (https://cpeditor.org)

#include <bits/stdc++.h>
#define FOR(i, a, b) for (int i = a; i <= b; ++i)
#define ROF(i, a, b) for (int i = a; i >= b; --i)
#define all(x) x.begin(),x.end()
#define debug(x) cerr<<x<<' '
#define CLR(i,a) memset(i,a,sizeof(i))

using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef pair<ll,ll> pll;
typedef array<ll,3> arr;
typedef double DB;
const ll MAXN=static_cast<ll>(1e3)+10,MAXval=static_cast<ll>(5e18)+3;

ll N,M,D;
char maze[MAXN][MAXN];
ll vis[MAXN][MAXN];

ll dx[4]={1,0,-1,0};
ll dy[4]={0,-1,0,1};

queue<arr> q;

inline void bfs(){	// 需要对之前所有加入q的点做好vis=true的处理
	while(!q.empty()){
		ll x=q.front()[0],y=q.front()[1],d=q.front()[2];
		q.pop();
		if(d>=D) continue;
		FOR(i,0,3){
			ll fx=x+dx[i],fy=y+dy[i];
			if(fx<1 || fx>N) continue;
			if(fy<1 || fy>M) continue;
			if(vis[fx][fy]<d+1) continue;
			if(maze[fx][fy]=='#') continue;
			vis[fx][fy]=d+1;
			q.push({fx,fy,d+1});
			// cerr<<fx<<' '<<fy<<'\n';
		}
	}
}

int main()
{
	ios::sync_with_stdio(false);
	cin.tie(0);cout.tie(0);
	
	cin>>N>>M>>D;
	FOR(i, 1, N){
		FOR(j,1,M){
			cin>>maze[i][j];
		}
	}
	CLR(vis,0x3f);
	ll stand=vis[0][0];
	FOR(i,1,N){
		FOR(j,1,M){
			if(maze[i][j]!='H') continue;
			q.push({i,j,0});
			vis[i][j]=0;
		}
	}
	bfs();
	ll ans=0;
	FOR(i,1,N){
		FOR(j,1,M){
			if(vis[i][j]!=stand) ++ans;
		}
	}
	cout<<ans<<endl;
	return 0;
}
/*
AC
https://atcoder.jp/contests/abc383/submissions/62366428
*/

心路历程（WA，TLE，MLE……）

Posted on Edited on

思路讲解

AC代码

https://atcoder.jp/contests/abc383/submissions/62366428

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// Problem: C - Humidifier 3
// Contest: AtCoder - Daiwa Securities Co. Ltd. Programming Contest 2024（AtCoder Beginner Contest 383）
// URL: https://atcoder.jp/contests/abc383/tasks/abc383_c
// Memory Limit: 1024 MB
// Time Limit: 2000 ms
// 
// Powered by CP Editor (https://cpeditor.org)

#include <bits/stdc++.h>
#define FOR(i, a, b) for (int i = a; i <= b; ++i)
#define ROF(i, a, b) for (int i = a; i >= b; --i)
#define all(x) x.begin(),x.end()
#define debug(x) cerr<<x<<' '
#define CLR(i,a) memset(i,a,sizeof(i))

using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef pair<ll,ll> pll;
typedef array<ll,3> arr;
typedef double DB;
const ll MAXN=static_cast<ll>(1e3)+10,MAXval=static_cast<ll>(5e18)+3;

ll N,M,D;
char maze[MAXN][MAXN];
ll vis[MAXN][MAXN];

ll dx[4]={1,0,-1,0};
ll dy[4]={0,-1,0,1};

queue<arr> q;

inline void bfs(){	// 需要对之前所有加入q的点做好vis=true的处理
	while(!q.empty()){
		ll x=q.front()[0],y=q.front()[1],d=q.front()[2];
		q.pop();
		if(d>=D) continue;
		FOR(i,0,3){
			ll fx=x+dx[i],fy=y+dy[i];
			if(fx<1 || fx>N) continue;
			if(fy<1 || fy>M) continue;
			if(vis[fx][fy]<d+1) continue;
			if(maze[fx][fy]=='#') continue;
			vis[fx][fy]=d+1;
			q.push({fx,fy,d+1});
			// cerr<<fx<<' '<<fy<<'\n';
		}
	}
}

int main()
{
	ios::sync_with_stdio(false);
	cin.tie(0);cout.tie(0);
	
	cin>>N>>M>>D;
	FOR(i, 1, N){
		FOR(j,1,M){
			cin>>maze[i][j];
		}
	}
	CLR(vis,0x3f);
	ll stand=vis[0][0];
	FOR(i,1,N){
		FOR(j,1,M){
			if(maze[i][j]!='H') continue;
			q.push({i,j,0});
			vis[i][j]=0;
		}
	}
	bfs();
	ll ans=0;
	FOR(i,1,N){
		FOR(j,1,M){
			if(vis[i][j]!=stand) ++ans;
		}
	}
	cout<<ans<<endl;
	return 0;
}
/*
AC
https://atcoder.jp/contests/abc383/submissions/62366428
*/

心路历程（WA，TLE，MLE……）

Posted on Edited on

思路讲解

参考洛谷题解

在kruskal的过程中求解这个问题，然后如果发现合并的复杂度太高，看看用数量代替set的合并可不可以？

然后记得结构体的全部数组要初始化

AC代码

https://atcoder.jp/contests/abc383/submissions/62364335

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// Problem: E - Sum of Max Matching
// Contest: AtCoder - Daiwa Securities Co. Ltd. Programming Contest 2024（AtCoder Beginner Contest 383）
// URL: https://atcoder.jp/contests/abc383/tasks/abc383_e
// Memory Limit: 1024 MB
// Time Limit: 2500 ms
// 
// Powered by CP Editor (https://cpeditor.org)

#include <bits/stdc++.h>
#define FOR(i, a, b) for (int i = a; i <= b; ++i)
#define ROF(i, a, b) for (int i = a; i >= b; --i)
#define all(x) x.begin(),x.end()
#define debug(x) cerr<<x<<'\n'

using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef pair<ll,ll> pll;
typedef array<ll,3> arr;
typedef double DB;
const ll MAXN=static_cast<ll>(2e5)+10,MAXval=static_cast<ll>(5e18)+3;

ll N,M,K;
vector<pll> G[MAXN];
struct Triple{
	ll u,v,w;
};
struct cmp{
	bool operator()(const Triple &a,const Triple &b)const{
		return a.w>b.w;
	}
};
priority_queue<Triple,vector<Triple>,cmp> pq;

struct FA{
	                 // 初始每个点的未匹配点数量
	ll fa[MAXN],size,numa[MAXN],numb[MAXN];
	inline void init(ll range){
		size=range;
		FOR(i,1,size){	// 初始化并查集
			fa[i]=i;
			numa[i]=0;
			numb[i]=0;
		}
	}
	ll find(ll x){
		if(fa[x]==x) return x;
		return fa[x]=find(fa[x]);
	}
	inline void merge(ll a,ll b){
		ll faA=find(a),faB=find(b);
		fa[faA]=faB;
		// 将a合并入b
		numa[faB]+=numa[faA];
		numb[faB]+=numb[faA];
		// debug(numa[faB]);
		// debug(numb[faB]);
	}
	inline ll match(ll a,ll b){
		ll faA=find(a),faB=find(b);
		ll aOption=min(numa[faA],numb[faB]),bOption=min(numb[faA],numa[faB]);
		numa[faA]-=aOption;
		numb[faB]-=aOption;
		numb[faA]-=bOption;
		numa[faB]-=bOption;
		return aOption+bOption;
	}
};

int main()
{
	ios::sync_with_stdio(false);
	cin.tie(0);cout.tie(0);
	cin>>N>>M>>K;
	FA fa;
	fa.init(N); 
	FOR(i, 1, M){
		ll u,v,w;
		cin>>u>>v>>w;
		G[u].push_back({v,w});
		G[v].push_back({u,w});
		pq.push((Triple){u,v,w});
	}
	FOR(i,1,K){
		ll a;
		cin>>a;
		fa.numa[a]+=1;
	}
	FOR(i,1,K){
		ll b;
		cin>>b;
		fa.numb[b]+=1;
	}
	// FOR(i,1,N){
		// cerr<<fa.numa[i]<<' ';
	// }
	// cerr<<'\n';
	// FOR(i,1,N){
		// cerr<<fa.numb[i]<<' ';
	// }
	// cerr<<'\n';
	ll ans=0;
	// kruskal（最小生成树+无向图判环）
	while(!pq.empty()){
		ll u=pq.top().u,v=pq.top().v,w=pq.top().w;
		pq.pop();
		if(fa.find(u)!=fa.find(v)){
			ans+=w*fa.match(u,v);
			// cerr<<u<<' '<<v<<' '<<w<<' '<<ans<<' '<<'\n';
			fa.merge(u,v);
		}
	}
	cout<<ans<<'\n';
	return 0;
}
/*

*/

心路历程（WA，TLE，MLE……）

https://atcoder.jp/contests/abc383/submissions/62364233

类的数组记得要init（）

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inline void init(ll range){
	size=range;
	FOR(i,1,size){	// 初始化并查集
		fa[i]=i;
		numa[i]=0;
		numb[i]=0;
	}
}

Posted on Edited on

题目大意

给定正整数 NN，求不超过 NN 的正整数中，恰好有 9 个正约数的数的个数。

输入格式：

  • 一行一个整数 NN （ 1≤N≤4×10121 \leq N \leq 4 \times 10^{12} ）

输出格式：

  • 一个整数，表示答案

样例说明：

  • 当 N=200N = 200 时，满足条件的正整数有 36, 100, 196 共 3 个

思路讲解

约数个数定理（因数个数定理）

这篇解释了该定理

image

AC代码

https://atcoder.jp/contests/abc383/submissions/62341051

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// Problem: D - 9 Divisors
// Contest: AtCoder - Daiwa Securities Co. Ltd. Programming Contest 2024（AtCoder Beginner Contest 383）
// URL: https://atcoder.jp/contests/abc383/tasks/abc383_d
// Memory Limit: 1024 MB
// Time Limit: 2000 ms
// 
// Powered by CP Editor (https://cpeditor.org)

#include <bits/stdc++.h>
#define FOR(i, a, b) for (int i = a; i <= b; ++i)
#define ROF(i, a, b) for (int i = a; i >= b; --i)
#define all(x) x.begin(),x.end()
#define deSpace(x) cerr<<x<<' '
#define deEnter(x) cerr<<x<<'\n'

using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef pair<ll,ll> pll;
typedef array<ll,3> arr;
typedef double DB;
const ll MAXN=static_cast<ll>(4e6)+10,MAXval=static_cast<ll>(5e18)+3;

ll N;
bool isprime[MAXN];
ll prime[MAXN];
inline void genPrimels(ll range){
	FOR(i,2,range){	// 0,1 比较特殊
		isprime[i]=true;
	}
	for(int i=2;i*i<=range;++i){
		for(int j=i*2;j<=range;j+=i){
			isprime[j]=false;
		}
	}
}

int main()
{
	ios::sync_with_stdio(false);
	cin.tie(0);cout.tie(0);
	cin>>N;
	ll ans=0;
	ll sqrtN=floor(sqrt(N));
	genPrimels(sqrtN);
	ll idx=0;
	FOR(i,2,sqrtN){
		if(isprime[i]) prime[++idx]=i;
	}
	// deEnter(idx);
	ll flag=idx;
	FOR(i,1,idx){
		ROF(j,flag,1){
			if(prime[i]*prime[j]<=sqrtN){
				flag=j;
				break;
			}
		}
		if(flag<=i) break;
		ans+=flag-i;
	}
	FOR(i,1,idx){
		if(prime[i]*prime[i]*prime[i]*prime[i]*prime[i]*prime[i]*prime[i]*prime[i]<=N){
			++ans;
		}else{
			break;
		}
	}
	cout<<ans<<'\n';
	return 0;
}
/*
AC
https://atcoder.jp/contests/abc383/submissions/62341051
*/

心路历程（WA，TLE，MLE……）

暴力，TLE

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// Problem: D - 9 Divisors
// Contest: AtCoder - Daiwa Securities Co. Ltd. Programming Contest 2024（AtCoder Beginner Contest 383）
// URL: https://atcoder.jp/contests/abc383/tasks/abc383_d
// Memory Limit: 1024 MB
// Time Limit: 2000 ms
// 
// Powered by CP Editor (https://cpeditor.org)

#include <bits/stdc++.h>
#define FOR(i, a, b) for (int i = a; i <= b; ++i)
#define ROF(i, a, b) for (int i = a; i >= b; --i)
#define all(x) x.begin(),x.end()
#define deSpace(x) cerr<<x<<' '
#define deEnter(x) cerr<<x<<'\n'

using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef pair<ll,ll> pll;
typedef array<ll,3> arr;
typedef double DB;
const ll MAXN=static_cast<ll>(1e6)+10,MAXval=static_cast<ll>(5e18)+3;

ll N;
inline bool check(ll x){
	ll res=0;
	for(int i=1;i*i<=x;++i){
		if(i*i==x){
			++res;
		}else if(x%i==0){
			res+=2;
		}
		if(res>9) return false;
	}
	if(res==9) return true;
	else return false;
}

int main()
{
	ios::sync_with_stdio(false);
	cin.tie(0);cout.tie(0);
	cin>>N;
	ll ans=0;
	for(int i=1;i*i<=N;++i){
		if(check(i*i)){
			++ans;
			deEnter(i*i);
		}
	}
	cout<<ans<<'\n';
	return 0;
}
/*

*/

Posted on Edited on

思路讲解

这个第一个贪心做法我觉得有点玄学，当然也敲了一遍，下面的第一个就是二分法

有些时候，你发现一个问题可以排序，这个时候就非常有可能是二分答案了

二分答案就是要剪枝，而且一旦发现rank ≥ m就立即返回，不要犹豫（也算是一种剪枝）

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inline bool check(ll x){
	ll rank=0;
	FOR(i,1,N){
		if(f(i,1,1)<x) break;
		FOR(j,1,N){
			if(f(i,j,1)<x) break;
			FOR(k,1,N){
				if(f(i,j,k)<x) break;
				++rank;
				if(rank>=M) return true;
			}
		}
	}
	// deEnter(rank);
	// if(rank>=M) return true;
	return false;
}

AC代码

https://atcoder.jp/contests/abc391/submissions/62332464

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#include <bits/stdc++.h>
#define FOR(i, a, b) for (int i = a; i <= b; ++i)
#define ROF(i, a, b) for (int i = a; i >= b; --i)
#define all(x) x.begin(),x.end()
#define deSpace(x) cerr<<x<<' '
#define deEnter(x) cerr<<x<<'\n'

using namespace std;

typedef unsigned long long ll;
typedef unsigned long long ull;
typedef pair<ll,ll> pll;
typedef array<ll,3> arr;
typedef double DB;
const ll MAXN=static_cast<ll>(2e5)+10,MAXval=static_cast<ll>(1e18)+3;

ll N,M,A[5][MAXN];
struct Triple{
	ll val,i,j,k;
};
struct cmp{
	bool operator()(const Triple &a,const Triple &b) const{
		return a.val<b.val;
	}
};
struct cmpSet{
	bool operator()(const Triple &a,const Triple &b) const{
		if(a.val!=b.val) return a.val<b.val;
		if(a.i!=b.i) return a.i<b.i;
		if(a.j!=b.j) return a.j<b.j;
		if(a.k!=b.k) return a.k<b.k;
		return false;
	}
};
priority_queue<Triple,vector<Triple> ,cmp> pq;
set<Triple,cmpSet> vis;

ll f(ll i,ll j,ll k){
	ll res=0;
	res=A[1][i]*A[2][j]+A[1][i]*A[3][k]+A[2][j]*A[3][k];
	return res;
}

int main()
{
	ios::sync_with_stdio(false);
	cin.tie(0);cout.tie(0);
	cin>>N>>M;
	FOR(i, 1, 3){
		FOR(j,1,N){
			cin>>A[i][j];
		}
		sort(&A[i][1],&A[i][N+1],greater<ll>());
	}
	ll rank=0;
	pq.push((Triple){f(1,1,1),1,1,1});
	while(!pq.empty()){
		++rank;
		// deSpace(pq.top().val);deSpace(pq.top().i);deSpace(pq.top().j);deEnter(pq.top().k);
		if(rank==M){
			cout<<pq.top().val<<'\n';
			return 0;
		}
		ll i=pq.top().i,j=pq.top().j,k=pq.top().k;
		pq.pop();
		if(vis.find((Triple){f(i+1,j,k),i+1,j,k})==vis.end() && i+1<=N){
			pq.push((Triple){f(i+1,j,k),i+1,j,k});
			vis.insert((Triple){f(i+1,j,k),i+1,j,k});
		}
		if(vis.find((Triple){f(i,j+1,k),i,j+1,k})==vis.end() && j+1<=N){
			pq.push((Triple){f(i,j+1,k),i,j+1,k});
			vis.insert((Triple){f(i,j+1,k),i,j+1,k});
		}
		if(vis.find((Triple){f(i,j,k+1),i,j,k+1})==vis.end() && k+1<=N){
			pq.push((Triple){f(i,j,k+1),i,j,k+1});
			vis.insert((Triple){f(i,j,k+1),i,j,k+1});
		}
		
	}
	return 0;
}
/*
AC
https://atcoder.jp/contests/abc391/submissions/62332464
*/

https://atcoder.jp/contests/abc391/submissions/62335145

二分法

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#include <bits/stdc++.h>
#define FOR(i, a, b) for (int i = a; i <= b; ++i)
#define ROF(i, a, b) for (int i = a; i >= b; --i)
#define all(x) x.begin(),x.end()
#define deSpace(x) cerr<<x<<' '
#define deEnter(x) cerr<<x<<'\n'

using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef pair<ll,ll> pll;
typedef array<ll,3> arr;
typedef double DB;
const ll MAXN=static_cast<ll>(2e5)+10,MAXval=static_cast<ll>(3e18)+3;

ll N,M,A[MAXN],B[MAXN],C[MAXN];

ll f(ll i,ll j,ll k){
	ll res=A[i]*B[j]+A[i]*C[k]+B[j]*C[k];
	return res;
}
inline bool check(ll x){
	ll rank=0;
	FOR(i,1,N){
		if(f(i,1,1)<x) break;
		FOR(j,1,N){
			if(f(i,j,1)<x) break;
			FOR(k,1,N){
				if(f(i,j,k)<x) break;
				++rank;
				if(rank>=M) return true;
			}
		}
	}
	// deEnter(rank);
	// if(rank>=M) return true;
	return false;
}

int main()
{
	ios::sync_with_stdio(false);
	cin.tie(0);cout.tie(0);
	cin>>N>>M;
	FOR(i, 1, N){
		cin>>A[i];
	}
	FOR(i, 1, N){
		cin>>B[i];
	}
	FOR(i, 1, N){
		cin>>C[i];
	}
	sort(&A[1],&A[N+1],greater<ll>());
	sort(&B[1],&B[N+1],greater<ll>());
	sort(&C[1],&C[N+1],greater<ll>());
	// 使用二分查找到正好有K(M)个数比它大的数（也就是第K个）
	ll l=0,r=MAXval;
	while(l<r){
		ll mid=l+r+1>>1;
		if(check(mid)){
			l=mid;
		}else{
			r=mid-1;
		}
	}
	cout<<l<<'\n';
	return 0;
}
/*
AC
https://atcoder.jp/contests/abc391/submissions/62335145
*/

心路历程（WA，TLE，MLE……）