这题一眼lct,然而

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
using namespace std;
const int maxn=100010;
int n,m,k,fa[maxn],sta1[maxn*50],sta2[maxn*50],top,tt,ans[maxn];
int getfa(int x){
    if(x!=fa[x]){
        sta1[++top]=x;sta2[top]=fa[x];
        fa[x]=getfa(fa[x]); 
    }
    return fa[x];
}
struct edg{
    int x,y,tim;
}e[maxn*2];
struct que{
    int num,c[5];
}q[maxn];
void cdq(int l,int r){
    int now=top;
    if(l==r){
        ans[l]=1;
        for(int i=1;i<=q[l].num;++i){
            int fx=getfa(e[q[l].c[i]].x);
            int fy=getfa(e[q[l].c[i]].y);
            if(fx!=fy){
                ans[l]=0;break;
            }
        }
        while(top!=now)fa[sta1[top]]=sta2[top],top--;
        return; 
    }
    ++tt;
    int mid=l+r>>1;
    for(int i=l;i<=mid;++i)
        for(int j=1;j<=q[i].num;++j){
            e[q[i].c[j]].tim=tt;
    }
    for(int i=mid+1;i<=r;++i)
        for(int j=1;j<=q[i].num;++j)
        if(e[q[i].c[j]].tim!=tt){
            int fx=getfa(e[q[i].c[j]].x);
            int fy=getfa(e[q[i].c[j]].y);
            if(fx!=fy){
                sta1[++top]=fx;sta2[top]=fa[fx];
                fa[fx]=fy;
            }
    }
    cdq(l,mid);
    while(top!=now){fa[sta1[top]]=sta2[top];top--;}
    ++tt;
    for(int i=mid+1;i<=r;++i)
        for(int j=1;j<=q[i].num;++j){
            e[q[i].c[j]].tim=tt;
        }
    for(int i=l;i<=mid;++i)
        for(int j=1;j<=q[i].num;++j)
        if(e[q[i].c[j]].tim!=tt){//把后面的所有边中前面没删的加上; 
            int fx=getfa(e[q[i].c[j]].x);
            int fy=getfa(e[q[i].c[j]].y);
            if(fx!=fy){
                sta1[++top]=fx;sta2[top]=fa[fx];
                fa[fx]=fy;
            }
    }
    cdq(mid+1,r);
} 
int main(){
    cin>>n>>m;
    for(int i=1;i<=n;++i)fa[i]=i;
    for(int i=1;i<=m;++i){
        scanf("%d%d",&e[i].x,&e[i].y);
        e[i].tim=0;
    }
    cin>>k;tt=1;
    for(int i=1;i<=k;++i){
        scanf("%d",&q[i].num);
        for(int j=1;j<=q[i].num;++j){
            scanf("%d",&q[i].c[j]);
            e[q[i].c[j]].tim=tt;
        }
    }
    for(int i=1;i<=m;++i)
    if(e[i].tim!=tt){
        int fx=getfa(e[i].x);
        int fy=getfa(e[i].y);
        if(fx!=fy)fa[fx]=fy;
    }
    cdq(1,k);
    for(int i=1;i<=k;++i){
        if(ans[i])puts("Connected");
        else puts("Disconnected");
    }
    return 0;
} 

 

题解说可以cdq+并查集,于是复习了一下cdq;

posted on 2018-03-22 16:29  湮灭之瞳  阅读(140)  评论(0编辑  收藏  举报