bzoj 4579 [Usaco2016 Open]Closing the Farm

4579: [Usaco2016 Open]Closing the Farm

Time Limit: 10 Sec  Memory Limit: 128 MB
Submit: 132  Solved: 80
[Submit][Status][Discuss]

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

Gold

【分析】

USACO签到题第二弹...

考虑反着加点，并查集维护当前连通块个数

【代码】

//bzoj 4579 #include<iostream>#include<cstring>#include<cstdio>#define ll long long#define M(a) memset(a,0,sizeof a)#define fo(i,j,k) for(i=j;i<=k;i++)using namespace std;const int mxn=200005;int n,m,cnt,num;bool vis[mxn];int cut[mxn],father[mxn],head[mxn],ans[mxn];struct edge {int next,to;} f[mxn<<1]; inline void add(int u,int v){f[++cnt].to=v,f[cnt].next=head[u],head[u]=cnt;}inline int read(){int x=0,f=1;char ch=getchar();while(ch<'0'||ch>'9') {if(ch=='-') f=-1;ch=getchar();}while(ch>='0'&&ch<='9') x=(x<<1)+(x<<3)+ch-'0',ch=getchar();return x*f;}inline int find(int x){if(x!=father[x]) father[x]=find(father[x]);return father[x];}int main(){int i,j,u,v,t1,t2;n=read(),m=read();fo(i,1,n) father[i]=i;fo(i,1,m){u=read(),v=read();add(u,v),add(v,u);}for(i=n;i>=1;i--) cut[i]=read();fo(i,1,n){u=cut[i],vis[u]=1,num++;for(int j=head[u];j;j=f[j].next){t1=find(u);v=f[j].to;if(!vis[v]) continue;t2=find(v);if(t1!=t2) father[t1]=t2,num--;}ans[i]=num;}for(i=n;i;i--)  if(ans[i]==1) printf("YES\n");  else printf("NO\n");return 0;}