雅礼集训2019 day7

Inverse

雅礼集训2019 day7_第1张图片
1 ≤ n ≤ 500 , 0 ≤ k ≤ 50 1 ≤ n ≤ 500,0 ≤ k ≤ 50 1n500,0k50, P是一个1到n的排列.
概率 d p dp dp好题。
思路:
定义 f i , j , k f_{i,j,k} fi,j,k表示 k k k轮变换之后 a p i > a p j a_{p_i}>a_{p_j} api>apj的概率。
然后考虑对当前轮翻转的区间 [ l , r ] [l,r] [l,r]分类转移。

  1. r < i    o r    l > j    o r    i < l < r < j r<i\ \ or\ \ l>j\ \ or\ \ i<l<r<j r<i  or  l>j  or  i<l<r<j,则 f i , j , k − 1 → f i , j , k f_{i,j,k-1}\rightarrow f_{i,j,k} fi,j,k1fi,j,k
  2. l ≤ i ≤ r < j l\le i\le r<j lir<j,则 f l + r − i , j , k − 1 → f i , j , k f_{l+r-i,j,k-1}\rightarrow f_{i,j,k} fl+ri,j,k1fi,j,k
  3. i < l ≤ j ≤ r i<l\le j\le r i<ljr,则 f i , l + r − j , k − 1 → f i , j , k f_{i,l+r-j,k-1}\rightarrow f_{i,j,k} fi,l+rj,k1fi,j,k
  4. l ≤ i < j ≤ r l\le i<j\le r li<jr,则 1 − f l + r − j , l + r − i → f i , j , k 1-f_{l+r-j,l+r-i}\rightarrow f_{i,j,k} 1fl+rj,l+rifi,j,k

利用二阶前缀和来优化转移可以做到 O ( n 2 k ) O(n^2k) O(n2k)
代码:

#include
#define ri register int
using namespace std;
const int rlen=1<<18|1;
inline char gc(){
	static char buf[rlen],*ib,*ob;
	(ib==ob)&&(ob=(ib=buf)+fread(buf,1,rlen,stdin));
	return ib==ob?-1:*ib++;
}
inline int read(){
	int ans=0;
	char ch=gc();
	while(!isdigit(ch))ch=gc();
	while(isdigit(ch))ans=((ans<<2)+ans<<1)+(ch^48),ch=gc();
	return ans;
}
typedef long long ll;
const int mod=1e9+7;
inline int add(const int&a,const int&b){return a+b>=mod?a+b-mod:a+b;}
inline int dec(const int&a,const int&b){return a>=b?a-b:a-b+mod;}
inline int mul(const int&a,const int&b){return (ll)a*b%mod;}
inline void update(int&a,const int&b){a=a+b>=mod?a+b-mod:a+b;}
inline int ksm(int a,int p){int ret=1;for(;p;p>>=1,a=mul(a,a))if(p&1)ret=mul(ret,a);return ret;}
const int N=505;
int n,k,a[N],f[2][N][N],coe[N][N],tmp=0,ans=0,inv;
int sl[2][N][N],ssl[2][N][N],sm[2][N][N],ssm[2][N][N],sr[2][N][N],ssr[2][N][N];
inline int calc(int x){return x*(x+1)/2;}
inline void init(){
	inv=ksm(calc(n),mod-2);
	for(ri v,i=1;i<n;++i)for(ri j=i+1;j<=n;++j){
		v=a[i]>a[j];
		coe[i][j]=add(calc(j-i-1),add(calc(i-1),calc(n-j)));
		ssl[tmp][i][j]=add(ssl[tmp][i-1][j],sl[tmp][i][j]=add(sl[tmp][i-1][j],v));
		ssr[tmp][i][j]=add(ssr[tmp][i][j-1],sr[tmp][i][j]=add(sr[tmp][i][j-1],v));	
		ssm[tmp][i][j-i]=add(ssm[tmp][i-1][j-i],sm[tmp][i][j-i]=add(sm[tmp][i-1][j-i],dec(1,v)));
		f[tmp][i][j]=v;
	}
}
int main(){
	n=read(),k=read();
	for(ri i=1;i<=n;++i)a[i]=read();
	init();
	while(k--){
		tmp^=1;
		for(ri val,i=1;i<n;++i)for(ri j=i+1;j<=n;++j){
			val=mul(f[tmp^1][i][j],coe[i][j]);
			update(val,dec(dec(ssl[tmp^1][j-1][j],ssl[tmp^1][j-i-1][j]),ssl[tmp^1][i-1][j]));
			update(val,dec(add(ssr[tmp^1][i][n],ssr[tmp^1][i][i-1]),add(ssr[tmp^1][i][i+n-j],ssr[tmp^1][i][j-1])));
			update(val,dec(dec(ssm[tmp^1][n+i-j][j-i],ssm[tmp^1][n-j][j-i]),ssm[tmp^1][i-1][j-i]));
			val=mul(val,inv);
			ssl[tmp][i][j]=add(ssl[tmp][i-1][j],sl[tmp][i][j]=add(sl[tmp][i-1][j],val));
			ssr[tmp][i][j]=add(ssr[tmp][i][j-1],sr[tmp][i][j]=add(sr[tmp][i][j-1],val));
			ssm[tmp][i][j-i]=add(ssm[tmp][i-1][j-i],sm[tmp][i][j-i]=add(sm[tmp][i-1][j-i],dec(1,val)));
			f[tmp][i][j]=val;
		}
	}
	for(ri i=1;i<n;++i)for(ri j=i+1;j<=n;++j)update(ans,f[tmp][i][j]);
	cout<<ans;
	return 0;
}

Subsequence

雅礼集训2019 day7_第2张图片
n ≤ 1 0 5 , ∣ A i ∣ ≤ 1 0 7 . n ≤ 10^5,|Ai|≤ 10^7. n105,Ai107.
思路:
先考虑 O ( n 2 ) d p O(n^2)dp O(n2)dp做法, f i , j f_{i,j} fi,j表示前 i i i个选 j j j个出来的最值。
那么显然 f i , j = max ⁡ { f i − 1 , j , f i − 1 , j − 1 + a i ∗ j } f_{i,j}=\max\{f_{i-1,j},f_{i-1,j-1}+a_i*j\} fi,j=max{fi1,j,fi1,j1+aij}
然后通过打表/证明发现对于每一个 i i i,转移方程中满足后一种决策更优的 j j j一定是一段后缀。
证明:
雅礼集训2019 day7_第3张图片
于是可以用平衡树来维护 f i f_{i} fi的差分数组,每次从 f i f_i fi转移到 f i + 1 f_{i+1} fi+1的时候直接将 a i a_i ai二分插入,然后将比它打的全部加上这个 a i a_i ai的贡献。
代码:

#include
#define ri register int
using namespace std;
const int rlen=1<<18|1;
inline char gc(){
	static char buf[rlen],*ib,*ob;
	(ib==ob)&&(ob=(ib=buf)+fread(buf,1,rlen,stdin));
	return ib==ob?-1:*ib++;
}
inline int read(){
	int ans=0;
	bool f=1;
	char ch=gc();
	while(!isdigit(ch))f^=ch=='-',ch=gc();
	while(isdigit(ch))ans=((ans<<2)+ans<<1)+(ch^48),ch=gc();
	return f?ans:-ans;
}
typedef long long ll;
const int N=1e5+5;
int n;
ll sum=0,a[N];
namespace Splay{
	#define lc (son[p][0])
	#define rc (son[p][1])
	int son[N][2],siz[N],fa[N],rt=0,tot=0;
	ll val[N],add[N];
	inline bool which(const int&x){return x==son[fa[x]][1];}
	inline void pushup(int p){siz[p]=siz[lc]+1+siz[rc];}
	inline void pushnow(int p,ll v){add[p]+=v,val[p]+=v;}
	inline void pushdown(int p){
		if(add[p]){
			if(lc)pushnow(lc,add[p]);
			if(rc)pushnow(rc,add[p]);
			add[p]=0ll;
		}
	}
	inline void rotate(int x){
		int y=fa[x],z=fa[y],t=which(x);
		if(z)pushdown(z),son[z][which(y)]=x;
		pushdown(x),pushdown(y);
		fa[x]=z,fa[y]=x,son[y][t]=son[x][t^1],son[x][t^1]=y;
		if(son[y][t])fa[son[y][t]]=y;
		pushup(y),pushup(x);
	}
	inline void splay(int x){while(fa[x]){if(fa[fa[x]])rotate(which(x)^which(fa[x])?x:fa[x]);rotate(x);}rt=x;}
	inline int build(ll v,int ft,int t){return val[++tot]=v,siz[tot]=1,fa[tot]=ft,son[tot][0]=son[tot][1]=0,ft?son[ft][t]=tot:0,tot;}
	inline int insert(ll v){
		int p=rt,ft=0,t=0;
		ll coe=0;
		while(p){
			pushdown(ft=p);
			if((coe+siz[lc]+1)*v>val[p])p=lc,t=0;
			else coe+=siz[lc]+1,p=rc,t=1;
		}
		return splay(p=build((coe+1)*v,ft,t)),p;
	}
	inline void solve(int p){
		pushdown(p);
		if(lc)solve(lc);
		cout<<(sum+=val[p])<<' ';
		if(rc)solve(rc);
	}
}
int main(){
	n=read();
	for(ri i=1;i<=n;++i)a[i]=read();
	for(ri p,i=1;i<=n;++i){
		p=Splay::insert(a[i]);
		if(Splay::son[p][1])Splay::pushnow(Splay::son[p][1],a[i]);
	}
	Splay::solve(Splay::rt);
	return 0;
}

Convex

雅礼集训2019 day7_第4张图片
4 ≤ n ≤ 2 × 1 0 6 , ∣ x i ∣ , ∣ y i ∣ ≤ 1 0 9 . 4 ≤ n ≤ 2×10^6,|xi|,|yi|≤ 10^9. 4n2×106,xi,yi109. 保证给出的多边形严格为凸包.
思路:
假设一根对角线将多边形分成了 S a ≤ S b S_a\le S_b SaSb两部分,然后它对答案的贡献就是 A l l _ a r e a − 2 ∗ S a All\_area-2*S_a All_area2Sa,这启发我们只用求 S a S_a Sa的和,然后用类似旋转卡壳的方法可以卡出每一个点对 ( l , r ) (l,r) (l,r)满足 线 l r 线lr 线lr左边的面积不超过右边的。
发现答案是这么个东西: ∑ l i m = l + 2 r ∑ i = l + 2 l i m ( p i − p l ) ∗ ( p i − 1 − p l ) \sum_{lim=l+2}^{r}\sum_{i=l+2}^{lim}(p_{i}-p_l)*(p_{i-1}-p_l) lim=l+2ri=l+2lim(pipl)(pi1pl)
由于叉积具有分配律因此可以展开这个式子的后面变成求 ∑ l i m = l + 2 r ∑ i = l + 2 l i m p i ∗ p i − 1 − p l ∗ p i − 1 − p i ∗ p l \sum_{lim=l+2}^{r}\sum_{i=l+2}^{lim}p_i*p_{i-1}-p_l*p_{i-1}-p_i*p_l lim=l+2ri=l+2limpipi1plpi1pipl
⇒ A n s l , r = ∑ i = l + 2 r ( r − i ) ( p i ∗ p i − 1 − p l ∗ p i − 1 − p i ∗ p l ) \Rightarrow Ans_{l,r}=\sum_{i=l+2}^{r}(r-i)(p_i*p_{i-1}-pl*p_{i-1}-p_i*p_l) Ansl,r=i=l+2r(ri)(pipi1plpi1pipl)
于是维护一下 ∑ p i ∗ p i − 1 \sum p_i*p_{i-1} pipi1 ∑ p i \sum p_i pi就可以了。
代码:

#include
#define ri register int
using namespace std;
const int rlen=1<<18|1;
inline char gc(){
	static char buf[rlen],*ib,*ob;
	(ib==ob)&&(ob=(ib=buf)+fread(buf,1,rlen,stdin));
	return ib==ob?-1:*ib++;
}
inline int read(){
	int ans=0;
	bool f=1;
	char ch=gc();
	while(!isdigit(ch))f^=ch=='-',ch=gc();
	while(isdigit(ch))ans=((ans<<2)+ans<<1)+(ch^48),ch=gc();
	return f?ans:-ans;
}
typedef long long ll;
const int mod=1e9+7;
inline int add(const int&a,const int&b){return a+b>=mod?a+b-mod:a+b;}
inline int dec(const int&a,const int&b){return a>=b?a-b:a-b+mod;}
inline int mul(const int&a,const int&b){return (ll)a*b%mod;}
inline void update(int&a,const int&b){a=a+b>=mod?a+b-mod:a+b;}
inline int ksm(int a,int p){int ret=1;for(;p;p>>=1,a=mul(a,a))if(p&1)ret=mul(ret,a);return ret;}
const int N=2e6+5;
int n,s1[N<<1],ss1[N<<1],all=0,ans=0;
ll area=0;
struct pot{
	int x,y;
	pot(int x=0,int y=0):x(x),y(y){}
	friend inline pot operator+(const pot&a,const pot&b){return pot(add(a.x,b.x),add(a.y,b.y));}
	friend inline pot operator-(const pot&a,const pot&b){return pot(dec(a.x,b.x),dec(a.y,b.y));}
	friend inline pot operator*(const pot&a,const int&b){return pot(mul(a.x,b),mul(a.y,b));}
	friend inline int operator^(const pot&a,const pot&b){return dec(mul(a.x,b.y),mul(a.y,b.x));}
}a[N<<1],b[N],s2[N<<1],ss2[N<<1];
inline void modify(int l,int r){
	if(r<l)r+=n;
	int S1=0,sum;
	pot S2,S3;
	S1=dec(dec(ss1[r-1],ss1[l]),mul(dec(s1[r-1],s1[l]),2*n-r+1));
	S2=(ss2[r]-ss2[l+1])-(s2[r]-s2[l+1])*(2*n-r);
	S3=(ss2[r-1]-ss2[l])-(s2[r-1]-s2[l])*(2*n-r+1);
	sum=dec(S1,add(S2^a[l],a[l]^S3));
	update(ans,dec(mul(all,r-l-1),mul(2,sum)));
}
inline ll calc(int i,int j,int k){return (ll)(b[i].x-b[k].x)*(b[j].y-b[k].y)-(ll)(b[j].x-b[k].x)*(b[i].y-b[k].y);}
int main(){
	n=read();
	for(ri i=1;i<=n;++i)a[i].x=((b[i].x=read())%mod+mod)%mod,a[i].y=((b[i].y=read())%mod+mod)%mod,a[n+i]=a[i];
	for(ri i=3;i<=n;++i)area+=calc(i,i-1,1),update(all,(a[i]-a[1])^(a[i-1]-a[1]));
	a[n*2+1]=a[1];
	for(ri i=1;i<=n*2;++i){
		s1[i]=add(s1[i-1],a[i+1]^a[i]);
		ss1[i]=add(ss1[i-1],mul(n*2-i+1,a[i+1]^a[i]));
		s2[i]=s2[i-1]+a[i];
		ss2[i]=ss2[i-1]+a[i]*(n*2-i+1);
	}
	ll sum=0;
	for(ri l=1,r=2;l<=n;++l){
		while(sum+calc(r==n?1:r+1,r,l)<=area/2)sum+=calc(r==n?1:r+1,r,l),r=r==n?1:r+1;
		modify(l,r),sum-=calc(r,l+1,l);
	}
	cout<<ans;
	return 0;
}

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