题解
和BZOJ4025挺像的
就是维护边权是时间的最大生成树
删边直接删
两点未联通时直接相连，两点联通则找两点间边权小的一条边删除即可
代码
#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define pdi pair<db,int>
#define mp make_pair
#define pb push_back
#define enter putchar('\n')
#define space putchar(' ')
#define eps 1e-8
#define mo 974711
#define MAXN 500005
//#define ivorysi
using namespace std;
typedef long long int64;
typedef double db;
template<class T>
void read(T &res) {
res = 0;char c = getchar();T f = 1;
while(c < '0' || c > '9') {
if(c == '-') f = -1;
c = getchar();
}
while(c >= '0' && c <= '9') {
res = res * 10 + c - '0';
c = getchar();
}
res *= f;
}
template<class T>
void out(T x) {
if(x < 0) {x = -x;putchar('-');}
if(x >= 10) {
out(x / 10);
}
putchar('0' + x % 10);
}
int N,M;
int id[MAXN],tot,pos[MAXN + 5005];
int op[MAXN],x[MAXN],y[MAXN];
int t[5005][5005];
namespace lct {
struct node {
int lc,rc,fa,val,minq;
bool rev;
}tr[MAXN * 2];
#define lc(u) tr[u].lc
#define rc(u) tr[u].rc
#define fa(u) tr[u].fa
#define val(u) tr[u].val
#define minq(u) tr[u].minq
#define rev(u) tr[u].rev
void Init() {
val(0) = minq(0) = 0x7fffffff;
for(int i = 1 ; i <= N ; ++i) val(i) = minq(i) = M + 2;
}
void reverse(int u) {
swap(lc(u),rc(u));
rev(u) ^= 1;
}
void pushdown(int u) {
if(rev(u)) {
reverse(lc(u));
reverse(rc(u));
rev(u) = 0;
}
}
void update(int u) {
minq(u) = val(u);
minq(u) = min(minq(u),minq(lc(u)));
minq(u) = min(minq(u),minq(rc(u)));
}
bool isRoot(int u) {
if(!fa(u)) return true;
else return rc(fa(u)) != u && lc(fa(u)) != u;
}
bool which(int u) {
return rc(fa(u)) == u;
}
void rotate(int u) {
int v = fa(u);
if(!isRoot(v)) {(v == lc(fa(v)) ? lc(fa(v)) : rc(fa(v))) = u;}
fa(u) = fa(v);fa(v) = u;
if(u == lc(v)) {lc(v) = rc(u);fa(rc(u)) = v;rc(u) = v;}
else {rc(v) = lc(u);fa(lc(u)) = v;lc(u) = v;}
update(v);
}
void Splay(int u) {
static int que[MAXN],qr;
qr = 0;int x;
for(x = u ; !isRoot(x) ; x = fa(x)) que[++qr] = x;
que[++qr] = x;
for(int i = qr ; i >= 1 ; --i) pushdown(que[i]);
while(!isRoot(u)) {
if(!isRoot(fa(u))) {
if(which(fa(u)) == which(u)) rotate(fa(u));
else rotate(u);
}
rotate(u);
}
update(u);
}
void Access(int u) {
for(int x = 0 ; u ; x = u , u = fa(u)) {
Splay(u);
rc(u) = x;
update(u);
}
}
void Makeroot(int u) {
Access(u);Splay(u);reverse(u);
}
void Link(int u,int v) {
Makeroot(u);Makeroot(v);Splay(v);fa(v) = u;
}
void Cut(int u,int v) {
Makeroot(u);Access(v);Splay(u);
if(rc(u) == v) {rc(u) = 0;fa(v) = 0;update(u);}
}
int dfs(int u) {
if(val(u) == minq(u)) return u;
pushdown(u);
if(minq(lc(u)) == minq(u)) return dfs(lc(u));
else return dfs(rc(u));
}
int Query(int u,int v) {
Makeroot(u);Access(v);Splay(u);
return dfs(u);
}
bool Connected(int u,int v) {
Makeroot(u);Access(v);Splay(u);
int p = u;
while(rc(p)) p = rc(p);
if(p == v) return true;
return false;
}
}
using lct::Link;
using lct::Cut;
using lct::Makeroot;
using lct::Query;
using lct::Connected;
using lct::tr;
void Init() {
read(N);read(M);
tot = N;
lct::Init();
for(int i = 1 ; i <= M ; ++i) {
read(op[i]);read(x[i]);read(y[i]);
if(op[i] == 0) {
id[i] = ++tot;pos[tot] = i;
t[x[i]][y[i]] = t[y[i]][x[i]] = id[i];
tr[tot].val = tr[tot].minq = M + 1;
}
if(op[i] == 1) {
int k = t[x[i]][y[i]];
tr[k].val = tr[k].minq = i;
id[i] = k;
}
}
}
void Solve() {
for(int i = 1 ; i <= M ; ++i) {
if(op[i] == 0) {
if(!Connected(x[i],y[i])) {
Link(x[i],id[i]);Link(id[i],y[i]);
}
else {
int t = Query(x[i],y[i]);
if(lct::tr[t].val < lct::tr[id[i]].val) {
Cut(t,x[pos[t]]);Cut(t,y[pos[t]]);
Link(id[i],x[i]);Link(id[i],y[i]);
}
}
}
else if(op[i] == 1) {
Cut(x[i],id[i]);Cut(y[i],id[i]);
}
else {
if(Connected(x[i],y[i])) puts("Y");
else puts("N");
}
}
}
int main() {
#ifdef ivorysi
freopen("f1.in","r",stdin);
#endif
Init();
Solve();
return 0;
}