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;   
}