题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=6200
题意：给个图，有2种操作，一种是加一条无向边，二是查询u,v之间必须有的边的条数，所谓必须有的边就是对于u,v必须通过这条边才能到达。
解法：一个很简单的想法，搞出图上的一颗树，然后剩下的边当成询问点队加到更新点集，每加入一个更新点对，直接把u，v区间的值置为0即可，查询就直接区间求和，可以直接树剖来维护，简单暴力，读入挂卡过。还有1个log的做法，可以用LCT维护(这个没写，口胡的)
#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e5+5;
typedef long long LL;
struct edge{
int to,next;
edge(){}
edge(int to,int next):to(to),next(next){}
}E[maxn*2];
struct FastIO
{
static const int S = 1310720;
int wpos;
char wbuf[S];
FastIO() : wpos(0) {}
inline int xchar()
{
static char buf[S];
static int len = 0, pos = 0;
if (pos == len)
pos = 0, len = fread(buf, 1, S, stdin);
if (pos == len) return -1;
return buf[pos ++];
}
inline int xuint()
{
int c = xchar(), x = 0;
while (c <= 32) c = xchar();
for (; '0' <= c && c <= '9'; c = xchar()) x = x * 10 + c - '0';
return x;
}
inline int xint()
{
int s = 1, c = xchar(), x = 0;
while (c <= 32) c = xchar();
if (c == '-') s = -1, c = xchar();
for (; '0' <= c && c <= '9'; c = xchar()) x = x * 10 + c - '0';
return x * s;
}
inline void xstring(char *s)
{
int c = xchar();
while (c <= 32) c = xchar();
for (; c > 32; c = xchar()) * s++ = c;
*s = 0;
}
inline void wchar(int x)
{
if (wpos == S) fwrite(wbuf, 1, S, stdout), wpos = 0;
wbuf[wpos ++] = x;
}
inline void wint(LL x)
{
if (x < 0) wchar('-'), x = -x;
char s[24];
int n = 0;
while (x || !n) s[n ++] = '0' + x % 10, x /= 10;
while (n--) wchar(s[n]);
}
inline void wstring(const char *s)
{
while (*s) wchar(*s++);
}
~FastIO()
{
if (wpos) fwrite(wbuf, 1, wpos, stdout), wpos = 0;
}
} io;
int n, m, head[maxn],edgecnt, tim;
int sz[maxn], top[maxn], son[maxn], dep[maxn];
int fa[maxn],tid[maxn];
void init(){
memset(head,-1,sizeof(head));
memset(son,-1,sizeof(son));
edgecnt=tim=0;
}
void add(int u,int v){
E[edgecnt].to=v,E[edgecnt].next=head[u],head[u]=edgecnt++;
}
void dfs1(int u, int father, int d){
dep[u]=d;
fa[u]=father;
sz[u]=1;
for(int i=head[u]; i+1; i=E[i].next){
int v=E[i].to;
if(v!=father){
dfs1(v,u,d+1);
sz[u]+=sz[v];
if(son[u]==-1||sz[v]>sz[son[u]]) son[u]=v;
}
}
}
void dfs2(int u, int tp){
top[u]=tp;
tid[u]=++tim;
if(son[u]==-1) return;
dfs2(son[u],tp);
for(int i=head[u];i+1;i=E[i].next){
int v=E[i].to;
if(v!=son[u]&&v!=fa[u])
dfs2(v,v);
}
}
namespace DSU{
int fa1[maxn];
void init1(){
for(int i=1; i<maxn; i++) fa1[i]=i;
}
int find_set(int x){
if(x==fa1[x]) return x;
else return fa1[x] = find_set(fa1[x]);
}
bool union_set(int x, int y){
x = find_set(x);
y = find_set(y);
if(x!=y){
fa1[x]=y;
return 1;
}
else{
return 0;
}
}
}
using namespace DSU;
namespace SegmentTree{
int sum[maxn<<2], lazy[maxn<<2];
void pushup(int rt){
sum[rt]=sum[rt*2]+sum[rt*2+1];
}
void pushdown(int rt){
if(lazy[rt]){
lazy[rt*2]=lazy[rt];
lazy[rt*2+1]=lazy[rt];
sum[rt*2]=sum[rt*2+1]=0;
lazy[rt]=0;
}
}
void build(int l, int r, int rt){
lazy[rt] = 0;
if(l == r){
sum[rt] = l!=1;
return;
}
int mid = (l+r)/2;
build(l,mid,rt*2);
build(mid+1,r,rt*2+1);
pushup(rt);
}
void update(int L, int R, int l, int r, int rt){
if(L<=l&&r<=R){
lazy[rt]=1;
sum[rt]=0;
return;
}
pushdown(rt);
int mid=(l+r)/2;
if(L<=mid) update(L,R,l,mid,rt*2);
if(mid<R) update(L,R,mid+1,r,rt*2+1);
pushup(rt);
}
int query(int L, int R, int l, int r, int rt){
if(L<=l&&r<=R) return sum[rt];
int mid=(l+r)/2;
pushdown(rt);
int ret=0;
if(L<=mid) ret+=query(L,R,l,mid,rt*2);
if(mid<R) ret+=query(L,R,mid+1,r,rt*2+1);
return ret;
}
void update(int u, int v){
while(top[u]!=top[v]){
if(dep[top[u]]<dep[top[v]]) swap(u,v);
update(tid[top[u]], tid[u], 1, tim, 1);
u = fa[top[u]];
}
if(u == v) return;
if(dep[u]<dep[v]) swap(u,v);
update(tid[v]+1,tid[u],1,tim,1);
}
int query(int u, int v){
int ret=0;
while(top[u]!=top[v]){
if(dep[top[u]]<dep[top[v]]) swap(u,v);
ret += query(tid[top[u]], tid[u], 1, tim, 1);
u = fa[top[u]];
}
if(u==v) return ret;
if(dep[u]<dep[v]) swap(u,v);
ret += query(tid[v]+1,tid[u],1,tim,1);
return ret;
}
}
using namespace SegmentTree;
int main()
{
int T,q,ks=0;
T = io.xint();
while(T--){
printf("Case #%d:\n", ++ks);
n = io.xint();
m = io.xint();
init();
init1();
vector <pair<int, int> > G;
for(int i=1; i<=m; i++){
int u, v;
u = io.xint();
v = io.xint();
if(union_set(u,v)){
add(u, v);
add(v, u);
}else{
G.emplace_back(u, v);
}
}
dfs1(1,0,0);
dfs2(1,1);
build(1,n,1);
for(auto &it:G){
update(it.first,it.second);
}
q = io.xint();
for(int i=1; i<=q; i++){
int op,x,y;
op = io.xint();
x = io.xint();
y = io.xint();
if(op==1)
update(x, y);
else{
printf("%d\n", query(x,y));
}
}
}
return 0;
}