[poi2007] biu

题意：给定一个图，点n<=10⁵,边m<=10⁶,现在求它的补图有多少个联通分量。。

思路：很容易想到并查集，但是补图边太多了。。

于是只能优化掉一些多余的边。。

具体做法是用队列优化。。

刚开始把所有没被连接的点用一个链表连接起来，那么选取一个未扩展得点u，就可以吧链表分成两个，一个就是跟他补图有边的，一个是没边的。。

并把有边的跟他连接起来。。删除链表。。并放入队列。。

这样每个点进入队列一次，每条边被访问2次。。时间复杂度为O(n+m)

code:

 #include<cstdio>
 #include<iostream>
 #include<cstring>
 #include<cstdlib>
 #include<cmath>
 #include<algorithm>
 #include<string>
 #include<map>
 #include<set>
 #include<vector>
 #include<queue>
 #include<stack>
 #include<ctime>
 #define repf(i, a, b) for (int i = (a); i <= (b); ++i)
 #define repd(i, a, b) for (int i = (a); i >= (b); --i)
 #define M0(x)  memset(x, 0, sizeof(x))
 #define Inf  0x7fffffff
 #define MP make_pair
 #define PB push_back
 #define eps 1e-8
 #define pi acos(-1.0)
 using namespace std;
 const int maxn = ;
 const int maxm = ;
 struct edge{
       int v, next;
 } e[maxm];
 struct node{
       node *x, *y;
 } n1[maxn], n2[maxn], *list[], *d[], tr, *null = &tr;
 int fa[maxn], last[maxn], tot, n, m;

 void add(const int& u, const int& v){
      e[tot] = (edge){v, last[u]}, last[u] = tot++;
 }

 void init(){
      memset(last, -, sizeof(last));
      int u, v;
      for (int i = ; i < m; ++i){
            scanf("%d%d", &u, &v);
            add(u, v), add(v, u);
      }
 }

 int sz[maxn], inlist[maxn];
 int find(int k){
     return fa[k] == k ? k : fa[k] = find(fa[k]);
 }

 void delnode(node *it){
      it->x->y = it->y, it->y->x = it->x;
      it->x = it->y = null;
 }

 void addnode(node *list, node *it){
      it->x = list, it->y = list->y;
      list->y->x = it, list->y = it;
 }

 void clear(int c){
      node *uu;
      for (node *it = list[c]; it != null;){
             uu = it->y;
             it->x = it->y = null;
             it = uu;
      }
 }

 void solve(){
      M0(n1), M0(n2), M0(inlist);
      list[] = n1 + n + , list[] = n2 + n + ;
      node *it, *uu;
      list[]->x = list[]->y = null, list[]->x = list[]->y = null;
      d[] = n1, d[] = n2;
      int c = , u, v;
      for (int i = ; i <= n; ++i) addnode(list[c], d[c] + i), inlist[i] = ;
      for (int i = ; i <= n; ++i) fa[i] = i;
      queue<int> q;
      while (list[c]->y != null){
          u = list[c]->y - d[c];
          q.push(u);
          while (!q.empty()){
              int s = c ^ ;
              u = q.front(), q.pop();
              clear(s);
              for (int p = last[u]; ~p; p = e[p].next){
                   v = e[p].v;
                   if (!inlist[v]) continue;
                   delnode(d[c] + v);
                   addnode(list[s], d[s] + v);
              }
              int fu = find(u), fv;
              for (it = list[c]->y; it != null; it = it->y){
                   v = it - d[c];
                   if (!inlist[v]) continue;
                   fv = find(v);
                   if (fu != fv) fa[fv] = fu;
                   inlist[v] = , q.push(v);
              }
              c = s;
          }
      }
      M0(sz);
      for (int i = ; i <= n; ++i)
          ++sz[find(i)];
      vector<int> ans;
      for (int i = ; i <= n; ++i) if (sz[i])
           ans.push_back(sz[i]);
      sort(ans.begin(), ans.end());
      printf("%d\n", (int)ans.size());
      for (int i = ; i < (int)ans.size(); ++i)
           i == ans.size()- ? printf("%d\n", ans[i]) : printf("%d ", ans[i]);

 }

 int main(){
 //    freopen("a.in", "r", stdin);
 //    freopen("a.out", "w", stdout);
     while (scanf("%d%d", &n, &m) != EOF){
            init();
            solve();
     }
     return ;
 }