LOJ#2888. 「APIO2015」巴邻旁之桥 Palembang Bridges 贪心+splay

脑残了，这题竟然都不会.

显然，把所有左右端点放在一起排序，然后取中位数是 $k=1$ 时最优的.

然后 $k=2$ 的话枚举中间的分割点，分割点左右就变成两个子问题了.

动态求中位数的话用平衡树/权值线段树维护就行了.

code: 

#include <bits/stdc++.h>   
#define ll long long   
#define lson s[x].ch[0] 
#define rson s[x].ch[1] 
#define N 200009   
#define setIO(s) freopen(s".in","r",stdin) 
using namespace std;         
int tot,rt1,rt2,n,K,cnt;  
ll seq[N<<1];        
struct data 
{ 
	ll sum,w; 
	int ch[2],f,si;      
}s[N<<2];     
inline int get(int x) { return s[s[x].f].ch[1]==x; }      
inline void pushup(int x) 
{  
	s[x].si=s[lson].si+s[rson].si+1;  	
	s[x].sum=s[lson].sum+s[rson].sum+s[x].w;   
}
void rotate(int x) 
{
	int old=s[x].f,fold=s[old].f,which=get(x);  
	s[old].ch[which]=s[x].ch[which^1];  
	if(s[old].ch[which]) s[s[old].ch[which]].f=old;  
	s[x].ch[which^1]=old,s[old].f=x,s[x].f=fold;  
	if(fold) s[fold].ch[s[fold].ch[1]==old]=x;  
	pushup(old),pushup(x);  
}
void splay(int x,int &tar) 
{
	int u=s[tar].f;  
	for(int fa;(fa=s[x].f)!=u;rotate(x)) 
	    if(s[fa].f!=u) rotate(get(fa)==get(x)?fa:x);       
	tar=x;  
}
int getkth(int x,int kth) 
{    
	while(1) 
	{
		if(s[lson].si+1==kth) break;  
		else if(kth<=s[lson].si) x=lson;  
		else kth-=(s[lson].si+1),x=rson;  
	}
	return x;  
}            
void ins(int &x,int fa,int v) 
{     
	if(!x) 
	{
		x=++tot,s[x].w=v,s[x].f=fa,pushup(x);  
		return;   
	}
	ins(s[x].ch[v>s[x].w],x,v),pushup(x);   
}          
int find(int x,int v) 
{
	while(1) 
	{
		if(s[x].w==v) break;  
		else x=s[x].ch[v>s[x].w];          
	}
	return x;   
} 
void del(int v) 
{                  
	int x=find(rt2,v),l,r;   
	splay(x,rt2),l=s[x].ch[0],r=s[x].ch[1];     
	if(!l) s[r].f=0,rt2=r;   
	else if(!r) s[l].f=0,rt2=l;  
	else 
	{
		while(s[l].ch[1]) l=s[l].ch[1]; 
		splay(l,s[x].ch[0]);    
		s[l].f=0,s[l].ch[1]=r,s[r].f=l,rt2=l,pushup(rt2);   
	} 
}    
void build(int &x,int fa,int l,int r) 
{ 
	int mid=(l+r)>>1;  
	s[x=++tot].f=fa;   
	s[x].w=seq[mid];        
	if(mid>l) build(lson,x,l,mid-1);  
	if(r>mid) build(rson,x,mid+1,r);               
	pushup(x);  
}
ll query(int &x) 
{
	int u=x; 
	int p=getkth(u,(s[u].si&1)?s[u].si/2+1:s[u].si/2);     
	splay(p,x);             
	ll ans=-s[lson].sum+(ll)s[x].w*s[lson].si-(ll)s[x].w*s[rson].si+s[rson].sum;   
	return ans;  
}               
struct node 
{
	ll l,r;   
	node(ll l=0,ll r=0):l(l),r(r){}   
	bool operator<(const node b) const { return (l+r)<(b.l+b.r);   }          
}nd[N];   
int main() 
{ 
    // setIO("input");  
	scanf("%d%d",&K,&n);         
	char a[2],b[2]; 
	ll ans=0,x,y,z; 
	for(int i=1;i<=n;++i) 
	{ 
		scanf("%s%lld%s%lld",a,&x,b,&y);     
		if(x>y) swap(x,y);   
		if(a[0]==b[0]) ans+=abs(y-x);   
		else ++ans,seq[++cnt]=x,seq[++cnt]=y,nd[cnt/2]=node(x,y);        
	}                 
	if(cnt==0) { printf("%lld
",ans); return 0; }    
	sort(nd+1,nd+1+(cnt/2));   
	sort(seq+1,seq+1+cnt),build(rt2,0,1,cnt);                
	if(K==1) { printf("%lld
",ans+query(rt2)); return 0; }               
	ll fin=100000000000000;                     
	for(int i=1;i<cnt/2;++i)            
	{       
		del(nd[i].l);    
		del(nd[i].r);             
		ins(rt1,0,nd[i].l); 
		if(i%7==0) splay(tot,rt1); 
		ins(rt1,0,nd[i].r); 
		if(i%7==0) splay(tot,rt1);     
		fin=min(fin,query(rt1)+query(rt2));    
	}
	printf("%lld
",fin+ans);  
	return 0; 
}