bzoj 4579: [Usaco2016 Open]Closing the Farm （并查集+离线）

4579: [Usaco2016 Open]Closing the Farm

Time Limit: 10 Sec  Memory Limit: 128 MB
Submit: 113  Solved: 69
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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

Gold

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题解：并查集+离线

删边不好实现，所以将询问离线，倒着向图中加边，然后用并查集维护连通性并记录集合中的元素个数，如果集合中的元素个数恰好等于此时图中的点数，就输出YES.

#include<iostream>#include<algorithm>#include<cstring>#include<cstdio>#include<cmath>#define N 400003using namespace std;int point[N],next[N],v[N],vis[N],n,m;int tot,a[N],pd[N],fa[N],sum[N];void add(int x,int y){tot++; next[tot]=point[x]; point[x]=tot; v[tot]=y;tot++; next[tot]=point[y]; point[y]=tot; v[tot]=x;}int find(int x){if (fa[x]==x) return x;fa[x]=find(fa[x]);return fa[x];}int main(){freopen("closing.in","r",stdin);freopen("closing.out","w",stdout);scanf("%d%d",&n,&m);for (int i=1;i<=m;i++){int x,y; scanf("%d%d",&x,&y);add(x,y);}for (int i=1;i<=n;i++) scanf("%d",&a[i]);for (int i=1;i<=n;i++) fa[i]=i,sum[i]=1;for (int i=n;i>=1;i--){vis[a[i]]=1; int x=a[i];  int r1=find(x);for (int j=point[x];j;j=next[j]) if (vis[v[j]]) { int r2=find(v[j]); if (r2!=r1) { fa[r2]=r1; sum[r1]+=sum[r2]; } }if (sum[r1]!=n-i+1) pd[i]=0;else pd[i]=1;}for (int i=1;i<=n;i++)  if (pd[i]) printf("YES\n"); else printf("NO\n");}