题目大意：多次询问保留编号在[L,R]之间的边图中的连通块数量。1e5。
题解：固定右端点向左扫等价于求最大生成树，那么每条边出现的时间是一个区间，维护这个区间的右端点+1（也就是其被挤掉的时间）即可。
#include<bits/stdc++.h>
#define rep(i,a,b) for(int i=a;i<=b;i++)
#define Rep(i,v) rep(i,0,(int)v.size()-1)
#define lint long long
#define ull unsigned lint
#define db double
#define pb push_back
#define mp make_pair
#define fir first
#define sec second
#define N 400010
#define INF (INT_MAX/10-10)
#define inf (INT_MIN/10+10)
#define gc getchar()
#define debug(x) cerr<<#x<<"="<<x
#define sp <<" "
#define ln <<endl
using namespace std;
typedef pair<int,int> pii;
typedef set<int>::iterator sit;
inline int inn()
{
int x,ch;while((ch=gc)<'0'||ch>'9');
x=ch^'0';while((ch=gc)>='0'&&ch<='9')
x=(x<<1)+(x<<3)+(ch^'0');return x;
}
int ch[N][2],s[N],val[N],fa[N],pf[N],rev[N],nxt[N],P[N],u[N],v[N];
inline int gw(int x) { return ch[fa[x]][1]==x; }
inline int push_up(int x) { return s[x]=min(min(s[ch[x][0]],s[ch[x][1]]),val[x]); }
inline int setc(int x,int y,int z)
{ if(!x) return fa[y]=0;ch[x][z]=y;if(y) fa[y]=x;return push_up(x); }
inline int rotate(int x)
{
int y=fa[x],z=fa[y],a=gw(x),b=gw(y),c=ch[x][a^1];
return swap(pf[x],pf[y]),setc(y,c,a),setc(x,y,a^1),setc(z,x,b);
}
inline int push_down(int x)
{
if(!rev[x]) return 0;
if(ch[x][0]) rev[ch[x][0]]^=1;
if(ch[x][1]) rev[ch[x][1]]^=1;
swap(ch[x][0],ch[x][1]),rev[x]=0;
return 0;
}
inline int all_down(int x) { return (fa[x]?all_down(fa[x]):0),(rev[x]?push_down(x):0); }
inline int splay(int x)
{ for(all_down(x);fa[x];rotate(x)) if(fa[fa[x]]) rotate(gw(x)==gw(fa[x])?fa[x]:x);return 0; }
inline int expose(int x)
{
splay(x);int y=ch[x][1];if(!y) return 0;
return pf[y]=x,ch[x][1]=fa[y]=0,push_up(x);
}
inline int splice(int x)
{
splay(x);int y=pf[x];if(!y) return 0;
return pf[x]=0,expose(y),splay(y),setc(y,x,1),1;
}
inline int access(int x) { expose(x);while(splice(x));return 0; }
inline int evert(int x) { return access(x),splay(x),rev[x]^=1; }
inline int link(int x,int y) { return evert(x),evert(y),splay(x),splay(y),pf[x]=y; }
inline int cut(int x,int y) { return evert(x),access(y),splay(x),ch[x][1]=fa[y]=0,push_up(x); }
inline int grt(int x) { while(fa[x]) x=fa[x];return x; }
inline int query(int x,int y) { evert(x),access(y),splay(x);return grt(y)==x?s[x]:0; }
struct segment{
int s;segment *ch[2];
}*T[N];
int build(segment* &rt,int l,int r)
{
rt=new segment;int mid=(l+r)>>1;if(l==r) return rt->s=0;
return build(rt->ch[0],l,mid),build(rt->ch[1],mid+1,r);
}
int update(segment* &x,segment* &y,int l,int r,int p)
{
x=new segment,x->ch[0]=y->ch[0],x->ch[1]=y->ch[1];
x->s=y->s+1;if(l==r) return 0;int mid=(l+r)>>1;
if(p<=mid) update(x->ch[0],y->ch[0],l,mid,p);
else update(x->ch[1],y->ch[1],mid+1,r,p);return 0;
}
int query(segment* &rt,int l,int r,int s,int t)
{
if(s<=l&&r<=t) return rt->s;int ans=0,mid=(l+r)>>1;
if(s<=mid) ans+=query(rt->ch[0],l,mid,s,t);
if(mid<t) ans+=query(rt->ch[1],mid+1,r,s,t);
return ans;
}
int query(int a,int b,int l,int r,int s,int t)
{ return query(T[b],l,r,s,t)-query(T[a-1],l,r,s,t); }
int main()
{
for(int Tcs=inn();Tcs;Tcs--)
{
int n=inn(),m=inn(),q=inn();
rep(i,1,m) u[i]=inn(),v[i]=inn(),nxt[i]=m+1,P[i]=i+n;
rep(i,1,m) val[P[i]]=s[P[i]]=i;rep(i,0,n) val[i]=s[i]=INF;
rep(i,1,n+m) ch[i][0]=ch[i][1]=fa[i]=pf[i]=rev[i]=0;
rep(i,1,m)
{
if(u[i]==v[i]) { nxt[i]=1;continue; }
int j=query(u[i],v[i]);
if(j) cut(P[j],u[j]),cut(P[j],v[j]),nxt[j]=i;
link(P[i],u[i]),link(P[i],v[i]);
}
// rep(i,1,m) debug(i)sp,debug(nxt[i])ln;
build(T[0],1,m+1);rep(i,1,m) update(T[i],T[i-1],1,m+1,nxt[i]);
for(int l,r;q;q--) l=inn(),r=inn(),printf("%d\n",n-query(l,r,1,m+1,r+1,m+1));
}
return 0;
}