BZOJ 1018: [SHOI2008]堵塞的交通traffic(线段树分治+并查集)

传送门

解题思路

  可以离线,然后确定每个边的出现时间,算这个排序即可。然后就可以线段树分治了,连通性用并查集维护,因为要撤销,所以要按秩合并,时间复杂度\(O(nlog^2 n)\)

代码

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<vector>

using namespace std;
const int N=100005;

inline int rd(){
	int x=0; char ch=getchar();
	while(!isdigit(ch)) ch=getchar();
	while(isdigit(ch)) x=(x<<1)+(x<<3)+ch-'0',ch=getchar();
	return x;	
}

int n,tot,ans[N],fa[N<<1],siz[N<<1],num,cnt;
struct Data{
	int x,y,op,id;	
	friend bool operator<(const Data A,const Data B){
		if(A.x==B.x && A.y==B.y) return A.id<B.id;
		if(A.x==B.x) return A.y<B.y;
		return A.x<B.x;
	}	
}tmp[N],ask[N];
struct Node{
	int x,y,t;	
	Node(int _x=0,int _y=0,int _t=0){
		x=_x; y=_y;	t=_t;
	}
};

inline int id(int x,int y){
	return (x-1)*n+y;	
}
inline int get(int x){
	while(x!=fa[x]) x=fa[x];
	return x;
}

struct Segment_Tree{
	vector<Node> v[N<<2];
	vector<Node> Ask[N];
	void update(int x,int l,int r,int L,int R,Node now){
		if(L<=l && r<=R){
			if(now.t) Ask[l].push_back(now);
			else v[x].push_back(now);
			return ;	
		}
		int mid=(l+r)>>1;
		if(L<=mid) update(x<<1,l,mid,L,R,now);
		if(mid<R) update(x<<1|1,mid+1,r,L,R,now);
	}
	void query(int x,int l,int r){
		int xx,yy,uu,vv; vector<Node> now;
		for(int i=0;i<v[x].size();i++){
			xx=v[x][i].x; yy=v[x][i].y;
			uu=get(xx); vv=get(yy);
			if(uu==vv) continue;
			if(siz[uu]<siz[vv]) swap(uu,vv);
			siz[uu]+=siz[vv]; fa[vv]=uu; 
			now.push_back(Node(uu,vv,0));
		}
		int mid=(l+r)>>1;
		if(l==r){
			for(int i=0;i<Ask[l].size();i++) {
				xx=get(Ask[l][i].x); yy=get(Ask[l][i].y);
				ans[Ask[l][i].t]=(xx==yy)?1:0;	
			}
		}
		else query(x<<1,l,mid),query(x<<1|1,mid+1,r);
		for(int i=0;i<now.size();i++){
			xx=now[i].x; yy=now[i].y;
			fa[yy]=yy; siz[xx]-=siz[yy];
		}	
	}	
}tree;

int main(){
	n=rd(); int x1,y1,x2,y2,id1,id2; char s[10];
	for(int i=1;i<=n*2;i++) fa[i]=i,siz[i]=1;
	while(1){
		scanf("%s",s+1);
		if(s[1]=='E') break; cnt++;
		x1=rd(),y1=rd(),x2=rd(),y2=rd();
		id1=id(x1,y1); id2=id(x2,y2);
		if(id1>id2) swap(id1,id2); 
		if(s[1]=='A') {
			num++; ask[num].id=cnt;
			ask[num].x=id1; ask[num].y=id2;	
			continue;
		}
		tot++; tmp[tot].id=cnt;
		tmp[tot].x=id1; tmp[tot].y=id2;
		if(s[1]=='O') tmp[tot].op=0;
		else tmp[tot].op=1;
	}
	sort(tmp+1,tmp+1+tot);
	for(int i=1;i<=num;i++) 
		tree.update(1,1,cnt,ask[i].id,ask[i].id,Node(ask[i].x,ask[i].y,i));
	for(int i=1;i<=tot;i++){
		if(tmp[i].op==0 && tmp[i+1].x==tmp[i].x && tmp[i+1].y==tmp[i].y)
			tree.update(1,1,cnt,tmp[i].id,tmp[i+1].id,Node(tmp[i].x,tmp[i].y,0)),++i;
		else tree.update(1,1,cnt,tmp[i].id,cnt,Node(tmp[i].x,tmp[i].y,0));
	}
	tree.query(1,1,cnt);
	for(int i=1;i<=num;i++)
		puts(ans[i]?"Y":"N");
	return 0;	
}
posted @ 2019-02-20 14:51  Monster_Qi  阅读(60)  评论(0编辑  收藏