题意：给一个无向图。每次查询破坏一条边，每次输出查询后连通图的个数。
思路：并查集。逆向思维，删边变成加边。
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
#include<iostream>
#define inf -100000000
#define LL long long
#define maxn 100005
using namespace std;
struct Edge
{
int x,y;
};
int parent[maxn];
int findset(int p)
{
return (parent[p]==p)?p:(parent[p]=findset(parent[p]));
}
Edge edge[maxn];
int query[maxn];
bool build[maxn];
int ans[maxn];
int main()
{
int n,m;
while(scanf("%d%d",&n,&m)!=EOF)
{
memset(build,,sizeof(build));
; i<=m; ++i)
scanf("%d%d",&edge[i].x,&edge[i].y);
int q;
scanf("%d",&q);
; i<=q; ++i)
{
scanf("%d",&query[i]);
build[query[i]]=true;
}
; i<=n; ++i)
parent[i]=i;
;
; i<=m; ++i)
if(!build[i])
{
int fa=findset(edge[i].x),fb=findset(edge[i].y);
if(fa!=fb)
parent[fa]=fb;
}
; i<=n; ++i)
if(findset(i)==i) cnt++;
; --i)
{
ans[i]=cnt;
int fa=findset(edge[query[i]].x),fb=findset(edge[query[i]].y);
if(fa!=fb)
{
parent[fa]=fb;
cnt--;
}
}
; i<=q; ++i)
) printf("%d",ans[i]);
else printf(" %d",ans[i]);
printf("\n");
}
;
}