bzoj4579: [Usaco2016 Open]Closing the Farm

bzoj4579: [Usaco2016 Open]Closing the Farm

Time Limit: 10 Sec  Memory Limit: 128 MB
Submit: 24  Solved: 17
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Description

Farmer John and his cows are planning to leave town for a long vacation, and so FJ wants to temporar

ily close down his farm to save money in the meantime.The farm consists of NN barns connected with M

M bidirectional paths between some pairs of barns (1≤N,M≤200,000). To shut the farm down, FJ plans

 to close one barn at a time. When a barn closes, all paths adjacent to that barn also close, and ca

n no longer be used.FJ is interested in knowing at each point in time (initially, and after each clo

sing) whether his farm is "fully connected" -- meaning that it is possible to travel from any open b

arn to any other open barn along an appropriate series of paths. Since FJ's farm is initially in som

ewhat in a state of disrepair, it may not even start out fully connected.

 

Input

The first line of input contains N and M. The next M lines each describe a path in terms of the pair

 of barns it connects (barns are conveniently numbered 1…N). The final N lines give a permutation o

f 1…N describing the order in which the barns will be closed.

 

Output

The output consists of N lines, each containing "YES" or "NO". The first line indicates whether the 

initial farm is fully connected, and line i+1 indicates whether the farm is fully connected after th

e iith closing.

 

Sample Input

4 3
1 2
2 3
3 4
3
4
1
2

Sample Output

YES
NO
YES
YES

HINT

 

Source

 直接在线的话lct似乎是可以做的？ 

 然而可以离线。。。所以就变成傻逼题了。。。

 离线后倒过来加边 并查集维护一下联通块个数即可

 

#include<bits/stdc++.h>
#define rep(i,l,r) for(int i=l;i<=r;i++)
#define N 500000
using namespace std;
struct E{
    int to,next;
}e[N];
int n,m,out[N],f[N],tot,s[N],flag[N],head[N];
inline int read() {
     register int x=0,ch=getchar();
     while(ch<'0'||ch>'9') ch=getchar();
     while(ch>='0'&&ch<='9') x=x*10+ch-48,ch=getchar();
     return x;
}
int find(int x) {
     if(f[x]==x) return x;else return f[x]=find(f[x]);
}
inline void ins(int u,int v) {
     e[++tot].to=v; e[tot].next=head[u]; head[u]=tot;
}
int main () {
     register int a,b,cnt,x,y;
     scanf("%d%d",&n,&m);
     rep(i,1,m) {a=read(); b=read(); ins(a,b); ins(b,a);}
     rep(i,1,n) s[i]=read(),flag[s[i]]=1,f[i]=i; cnt=n;
     rep(i,1,n) if(!flag[i]){
          for(register int k=head[i];k;k=e[k].next) {
               if(!flag[e[k].to]) {
                    x=find(i); y=find(e[k].to);
                    if(x!=y) f[x]=y,--cnt;
               }
          }
     }
     for(register int i=n;i;i--) {
          flag[s[i]]=0;
          for(register int k=head[s[i]];k;k=e[k].next) if(!flag[e[k].to]) {
               x=find(s[i]); y=find(e[k].to);
               if(x!=y) f[x]=y,--cnt;
          }
          out[i]=(cnt==i);
     }
     rep(i,1,n) if(!out[i]) puts("NO");else puts("YES");
}