bzoj 4579: [Usaco2016 Open]Closing the Farm

4579: [Usaco2016 Open]Closing the Farm

Description

Farmer John and his cows are planning to leave town for a long vacation, and so FJ wants to temporarily close down his farm to save money in the meantime.The farm consists of NN barns connected with MM 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 can no longer be used.FJ is interested in knowing at each point in time (initially, and after each closing) whether his farm is "fully connected" -- meaning that it is possible to travel from any open barn to any other open barn along an appropriate series of paths. Since FJ's farm is initially in somewhat 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

题解：

首先这道题目什么意思呢？？？

大概是讲用m条边连接的农场，每次关闭一个谷仓，问你每次在关闭谷仓之前整个农场是否完全联通。

知道意思后这就是到很裸的并查集了，维护联通块个数，看看是不是1就行了。。。

#include<stdio.h>
#include<iostream>
using namespace std;
const int N=200005;
int n,m,i,j,x,y,fx,fy,cnt,a[N],p[N],ans[N],f[N];
int tot,head[N],Next[N<<1],to[N<<1];
void add(int x,int y)
{
    tot++;
    to[tot]=y;
    Next[tot]=head[x];
    head[x]=tot;
}
int get(int x)
{
    if(f[x]==x) return x;else return f[x]=get(f[x]);
}
int main()
{
    scanf("%d%d",&n,&m);
    for(i=1;i<=n;i++)
        head[i]=-1;
    for(i=1;i<=m;i++)
    {
        scanf("%d%d",&x,&y);
        add(x,y);
        add(y,x);
    }
    for(i=1;i<=n;i++)
        f[i]=i;
    for(i=1;i<=n;i++)
        scanf("%d",&a[i]);
    cnt=0;
    for(i=n;i>=1;i--)
    {
        cnt++;
        x=a[i];
        for(j=head[x];j!=-1;j=Next[j])
        {
            y=to[j];
            if(p[y]==1)
            {
                fx=get(x);
                fy=get(y);
                if(fx!=fy)
                {
                    cnt--;
                    f[fx]=fy;
                }
            }
        }
        ans[i]=cnt;
        p[x]=1;
    }
    for(i=1;i<=n;i++)
        if(ans[i]==1) printf("YES\n");else printf("NO\n");
    return 0;
}