题目描述
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.
输入
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 of 1…N describing the order in which the barns will be closed.
输出
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 the iith closing.
样例输入
4 3
1 2
2 3
3 4
3
4
1
2
样例输出
YES
NO
YES
YES
题目大意
给你n个点和m条边的无向图，有n次删点操作，删掉点后与这个点相连的边也随之删除。问删除每个点之前这个图是不是连通图。
题解
并查集
由于删点比较难搞，所以我们需要换一种思路：
可以先把所有的点删掉，然后反过来一个一个再加进来。
这样便于直接处理改动的边。
然后用一个并查集维护连通块即可。
#include <cstdio>
int head[200010] , to[400010] , next[400010] , cnt , a[200010] , f[200010] , ans[200010] , ok[200010];
int find(int x)
{
return x == f[x] ? x : f[x] = find(f[x]);
}
void add(int x , int y)
{
to[++cnt] = y;
next[cnt] = head[x];
head[x] = cnt;
}
int main()
{
int n , m , i , j , x , y , tmp = 0;
scanf("%d%d" , &n , &m);
for(i = 1 ; i <= m ; i ++ )
scanf("%d%d" , &x , &y) , add(x , y) , add(y , x);
for(i = 1 ; i <= n ; i ++ )
scanf("%d" , &a[i]);
for(i = 1 ; i <= n ; i ++ )
f[i] = i;
for(i = n ; i >= 1 ; i -- )
{
ok[a[i]] = 1;
tmp ++ ;
for(j = head[a[i]] ; j ; j = next[j])
{
if(ok[to[j]])
{
x = find(a[i]) , y = find(to[j]);
if(x != y)
{
f[x] = y;
tmp -- ;
}
}
}
ans[i] = (tmp == 1);
}
for(i = 1 ; i <= n ; i ++ )
printf("%s
" , ans[i] ? "YES" : "NO");
return 0;
}