并查集好合并不好拆开 可以考虑离线 先读入 从大到小排序 再依次合并

技巧：不好断开就倒着来合并 JSOI2008 P1197 也是该思想

#include
     
      using namespace std;#define Max(x,y) (x)<(y)?(y):(x)#define Min(x,y) (x)<(y)?(x):(y)#define ll long long#define rg registerconst int N=300000+5,M=1000000+5,inf=0x3f3f3f3f,P=9999973;int n,m,f[N],sz[N];template 
      
       void rd(t &x){    x=0;int w=0;char ch=0;    while(!isdigit(ch)) w|=ch=='-',ch=getchar();    while(isdigit(ch)) x=(x<<1)+(x<<3)+(ch^48),ch=getchar();    x=w?-x:x;}int find(int x) {
   return f[x]==x?x:f[x]=find(f[x]);}void merge(int x,int y){    int fx=find(x),fy=find(y);    if(fx==fy) return;    sz[fx]+=sz[fy],f[fy]=fx;}struct edge{
   int u,v,w;}e[N<<1];struct node{
   int k,v,pos,ans;}ask[N];bool cmp1(edge a,edge b){
   return a.w>b.w;}bool cmp2(node a,node b){
   return a.k>b.k;}bool cmp3(node a,node b){
   return a.pos
       
        =ask[i].k)        merge(e[nxte].u,e[nxte].v),++nxte;        ask[i].ans=sz[find(ask[i].v)]-1;    }//    for(int i=1;i<=n;++i) printf("%d ",sz[i]);    sort(ask+1,ask+m+1,cmp3);    for(int i=1;i<=m;++i) printf("%d\n",ask[i].ans);    return 0;}