集訓 4 & 模擬 5

xrlong發表於2024-05-04

集訓 4 & 模擬 5

有點唐簡單，所以一起寫了（其實是因為之前懶得寫）

集訓 4：

T1 模擬，賽時不刪除錯保齡了。

T2 顯然貪心

T3 發現顯然要兩兩互質，有因為父比子小，所以方案數就是將 \(\varphi\) 乘起來（甚至都不需要線性篩）

T4 meet in middle 板子。

模擬 5

  1. T1 特殊字串

    顯然有 \(n^2\) 的 \(dp\)，發現字符集為 \(26\) 可以對結尾字元分別儲存 \(\max\) 最佳化到 \(n*26\)

    CODE 
    #include<bits/stdc++.h>
#include<sys/mman.h>
#include<fcntl.h>
using namespace std;
using llt=long long;
using llf=long double;
using ull=unsigned long long;
#define Ct const
#define Il __always_inline
#define For(i,a,b,c) for(int i=(a);i<=(b);i+=(c))
#define For_(i,a,b,c) for(int i=(a);i>=(b);i-=(c))
#define For_it(i,a,b) for(auto i=(a);i!=(b);++i)

namespace IO{
#ifdef ONLINE_JUDGE
	int Fin=open("shiki.in",0); FILE* Fout=fopen("shiki.out","w");
#elif defined(UN_FAST)
	FILE *Fin=freopen("in_out/in.in","r",stdin),*Fout=freopen("in_out/out.out","w",stdout);
#else
	int Fin=open("in_out/in.in",0); FILE* Fout=fopen("in_out/out.out","w");
#endif // file
#ifdef UN_FAST
	char cc;
	#define G (cc=getchar())
	#define C cc
#else
	const char *I=(char*)mmap(0,1<<28,1,2,Fin,0)-1; 
	#define G (*++I)
	#define C (*I)
#endif // fast (mmap)
#define P(x) putc_unlocked(x,Fout)
	template<class T> Il void read(T &x){x=0;bool f=0; while(f|=G==45,C<48); while(x=x*10+(C&15),G>47); f?x=-x:0;}
	template<class T> void write(T x){if(x<0)P('-'),x=-x; if(x/10)write(x/10); P('0'+x%10);}
	Il void read(char &c){while(G<33); c=C;} void read(char *s){int p=-1; while(G<33); while(s[++p]=C,G>32); s[++p]='\0';}
	Il void read(string &s){s.clear(); while(G<33); while(s.push_back(C),G>32);}
	template<class T=int> Il T read(){T a; read(a); return a;}
	template<class T,class... Argc> Il void read(T &x,Argc&... argc){read(x),read(argc...);}
	Il void write(const char &c){P(c);} Il void write(const char *s){for(const char *c=s;*c!='\0';++c) P(*c);}
	Il void write(const string &s){for(const char& c:s) P(c);}
	template<class T,class... Argc> Il void write(const T &x,const Argc&... argc){write(x),write(argc...);}
	template<class T> Il void Write(const T &x){write(x),P(' ');} Il void Write(const char &c){P(c); if(c>32)P(' ');}
	template<class T,class... Argc> Il void Write(const T &x,const Argc&... argc){Write(x),Write(argc...);}
#undef G
#undef C
#undef P
#define endl '\n'
}using IO::read; using IO::write; using IO::Write;

constexpr int N=1e5+3;
llt ma[30],n,m,wt[30][30],ans;
string s;

int main(){
	read(n,s,m);
	For(i,1,m,1){
		char a,b; int k; read(a,b,k);
		wt[a-'a'][b-'a']+=k;
	}
	memset(ma,-0x3f,sizeof(ma));
	ma[s[0]-'a']=0;
	For(i,2,n,1){
		int nw=s[i-1]-'a';
		For(j,0,25,1) ans=max(ans,ma[j]+wt[j][nw]);
		ma[nw]=max(ma[nw],ans);
	}
	write(ans);
}
// For(i,1,n,1) For(j,1,i-1,1) dp[i]=max(dp[i],dp[j]+wt[s[j-1]-'a'][s[i-1]-'a']);

  2. T2 寶可夢

    發現從一個點到自己的路徑是一個環，並且每條詢問路徑都是環上一段，直接預處理環即可（不太好寫，賽時想假了好幾次）。

    CODE 
    #include<bits/stdc++.h>
#include<sys/mman.h>
#include<fcntl.h>
using namespace std;
using llt=long long;
using llf=long double;
using ull=unsigned long long;
#define Ct const
#define Il __always_inline
#define For(i,a,b,c) for(int i=(a);i<=(b);i+=(c))
#define For_(i,a,b,c) for(int i=(a);i>=(b);i-=(c))
#define For_it(i,a,b) for(auto i=(a);i!=(b);++i)

namespace IO{
#ifdef ONLINE_JUDGE
	int Fin=open("pokemon.in",0); FILE* Fout=fopen("pokemon.out","w");
#elif defined(UN_FAST)
	FILE *Fin=freopen("in_out/in.in","r",stdin),*Fout=freopen("in_out/out.out","w",stdout);
#else
	int Fin=open("in_out/in.in",0); FILE* Fout=fopen("in_out/out.out","w");
#endif // file
#ifdef UN_FAST
	char cc;
	#define G (cc=getchar())
	#define C cc
#else
	const char *I=(char*)mmap(0,1<<28,1,2,Fin,0)-1; 
	#define G (*++I)
	#define C (*I)
#endif // fast (mmap)
#define P(x) putc_unlocked(x,Fout)
	template<class T> Il void read(T &x){x=0;bool f=0; while(f|=G==45,C<48); while(x=x*10+(C&15),G>47); f?x=-x:0;}
	template<class T> void write(T x){if(x<0)P('-'),x=-x; if(x/10)write(x/10); P('0'+x%10);}
	Il void read(char &c){while(G<33); c=C;} void read(char *s){int p=-1; while(G<33); while(s[++p]=C,G>32); s[++p]='\0';}
	Il void read(string &s){s.clear(); while(G<33); while(s.push_back(C),G>32);}
	template<class T=int> Il T read(){T a; read(a); return a;}
	template<class T,class... Argc> Il void read(T &x,Argc&... argc){read(x),read(argc...);}
	Il void write(const char &c){P(c);} Il void write(const char *s){for(const char *c=s;*c!='\0';++c) P(*c);}
	Il void write(const string &s){for(const char& c:s) P(c);}
	template<class T,class... Argc> Il void write(const T &x,const Argc&... argc){write(x),write(argc...);}
	template<class T> Il void Write(const T &x){write(x),P(' ');} Il void Write(const char &c){P(c); if(c>32)P(' ');}
	template<class T,class... Argc> Il void Write(const T &x,const Argc&... argc){Write(x),Write(argc...);}
#undef G
#undef C
#undef P
#define endl '\n'
}using IO::read; using IO::write; using IO::Write;

constexpr int N=1e5+3,NM=5e5+3,mx[]={0,0,-1,0,1},my[]={0,1,0,-1,0},tlt[]={0,2,3,4,1},trt[]={0,4,1,2,3};
vector<bool> c[N];
vector<int> pos[5][N];
int n,m,q,cnt;

void Dfs(Ct int &x,Ct int &y,Ct int &fw){
	if(pos[fw][x][y]) return ;
	pos[fw][x][y]=++cnt;
	int nf=trt[fw];
	For(i,1,4,1){
		int nx=x+mx[nf],ny=y+my[nf];
		if(nx>=1&&nx<=n&&ny>=1&&ny<=m&&c[nx][ny]) Dfs(nx,ny,nf);
		nf=tlt[nf];
	}
}
Il void Init(){
	int x,y;
	read(n,m);
	For(i,1,n,1){
		c[i].resize(m+3);
		For(k,1,4,1) pos[k][i].resize(m+3);
		For(j,1,m,1) if(read<char>()=='.') c[i][j]=1,x=i,y=j;
	}
	Dfs(x,y,2);
}

Il int Get(Ct int &xa,Ct int &ya,Ct int &fa,Ct int &xb,Ct int &yb,Ct int &fb){
	if(!pos[fa][xa][ya]||!pos[fb][xb][yb]) return 0x3f3f3f3f;
	if(pos[fa][xa][ya]<=pos[fb][xb][yb]) return pos[fb][xb][yb]-pos[fa][xa][ya];
	return cnt+pos[fb][xb][yb]-pos[fa][xa][ya]-1;
}

int main(){
	Init();
	read(q);
	For(i,1,q,1){
		int xa,ya,xb,yb,fa; char c; read(xa,ya,xb,yb,c);
		if(c=='R') fa=1; else if(c=='U') fa=2; else if(c=='L') fa=3; else fa=4;
		if(xa==xb&&ya==yb){write(0,'\n'); continue;}
		xa+=mx[fa],ya+=my[fa];
		int ans=0x3f3f3f3f; For(i,1,4,1) ans=min(ans,Get(xa,ya,fa,xb,yb,i)+1);
		write(ans,'\n');
	}
}

  3. T3 矩陣

    如果有相鄰且相同的，就是 \(-1\)

    其他情況對公比分治，在公比 \(>200\) 時最多長度為 \(2\)，可以直接整。

    公比 \(\le 200\) \(n^2*\text{公比}\) 暴力 \(dp\) 即可。

    也可以直接最長路。

    ps:

    誰 tm 造的資料，每個 sub 裡塞一個 ans=1 ，給我卡沒了。

    CODE 
    #include<bits/stdc++.h>
#include<sys/mman.h>
#include<fcntl.h>
using namespace std;
using llt=long long;
using llf=long double;
using ull=unsigned long long;
#define Ct const
#define Il __always_inline
#define For(i,a,b,c) for(int i=(a);i<=(b);i+=(c))
#define For_(i,a,b,c) for(int i=(a);i>=(b);i-=(c))
#define For_it(i,a,b) for(auto i=(a);i!=(b);++i)

namespace IO{
#ifdef ONLINE_JUDGE
	int Fin=open("matrix.in",0); FILE* Fout=fopen("matrix.out","w");
#elif defined(UN_FAST)
	FILE *Fin=freopen("in_out/in.in","r",stdin),*Fout=freopen("in_out/out.out","w",stdout);
#else
	int Fin=open("in_out/in.in",0); FILE* Fout=fopen("in_out/out.out","w");
#endif // file
#ifdef UN_FAST
	char cc;
	#define G (cc=getchar())
	#define C cc
#else
	const char *I=(char*)mmap(0,1<<28,1,2,Fin,0)-1; 
	#define G (*++I)
	#define C (*I)
#endif // fast (mmap)
#define P(x) putc_unlocked(x,Fout)
	template<class T> Il void read(T &x){x=0;bool f=0; while(f|=G==45,C<48); while(x=x*10+(C&15),G>47); f?x=-x:0;}
	template<class T> void write(T x){if(x<0)P('-'),x=-x; if(x/10)write(x/10); P('0'+x%10);}
	Il void read(char &c){while(G<33); c=C;} void read(char *s){int p=-1; while(G<33); while(s[++p]=C,G>32); s[++p]='\0';}
	Il void read(string &s){s.clear(); while(G<33); while(s.push_back(C),G>32);}
	template<class T=int> Il T read(){T a; read(a); return a;}
	template<class T,class... Argc> Il void read(T &x,Argc&... argc){read(x),read(argc...);}
	Il void write(const char &c){P(c);} Il void write(const char *s){for(const char *c=s;*c!='\0';++c) P(*c);}
	Il void write(const string &s){for(const char& c:s) P(c);}
	template<class T,class... Argc> Il void write(const T &x,const Argc&... argc){write(x),write(argc...);}
	template<class T> Il void Write(const T &x){write(x),P(' ');} Il void Write(const char &c){P(c); if(c>32)P(' ');}
	template<class T,class... Argc> Il void Write(const T &x,const Argc&... argc){Write(x),Write(argc...);}
#undef G
#undef C
#undef P
#define endl '\n'
}using IO::read; using IO::write; using IO::Write;

constexpr int N=4e4+3,mx[]={0,1,-1,0,0},my[]={0,0,0,1,-1};
vector<int> c[N],dp[N];
int n,m,ans=1;
int dfs(int x,int y,int k){
	if(dp[x][y]) return dp[x][y];
	else {
		For(i,1,4,1){
			int nx=x+mx[i],ny=y+my[i];
			if(nx>n||nx<1||ny>m||ny<1||c[nx][ny]!=c[x][y]*k) continue;
			dp[x][y]=max(dp[x][y],dfs(nx,ny,k));
		}
		dp[x][y]+=1;
		return dp[x][y];
	}
}

int main(){
	For(i,1,N-3,1) c[i].push_back(0);
	read(n,m);
	For(i,1,m,1){
		c[1].push_back(read());
		if(c[1][i]==c[1][i-1]){write(-1); return 0;}
		else if(i!=1&&(c[1][i]%c[1][i-1]==0||c[1][i-1]%c[1][i]==0)) ans=2;
	}
	For(i,2,n,1) For(j,1,m,1){
		c[i].push_back(read());
		if(c[i][j]==c[i][j-1]||c[i][j]==c[i-1][j]){write(-1); return 0;}
		else if(j!=1&&(c[i][j]%c[i][j-1]==0||c[i][j-1]%c[i][j]==0||c[i][j]%c[i-1][j]==0||c[i-1][j]%c[i][j]==0)) ans=2;
	}
	For(i,2,200,1){
		For(j,1,n,1) dp[j].clear(),dp[j].resize(m+1);
		For(a,1,n,1) For(b,1,m,1) ans=max(ans,dfs(a,b,i));
	}
	write(ans);
}

  4. T4 乘法

    這是真唐，整個拉插板子，我也不會，題解也不會 😅😅😅

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