题解 Dove 打扑克
考场上觉得复杂度是假的就没怎么优化，然后考完题解帮我证明了它是真的……
首先合并可以用并查集维护，可以顺便维护出集合的大小
对于操作2，发现如果 \(size_i\) 是确定的，可以用权值线段树很方便的维护出合法的 \(size_j\)的个数
每次只需枚举出现过的 \(size_i\) 即可，所以我觉得复杂度是假的
- 如果存在 \(\sum a_i = n\) ，那么有 \(diff \{a_i\} \leqslant \sqrt n\)
我们有 \(\sum size_i = n\) ，所以不同的size个数只有不超过根号个
所以用个unordered_set维护当前存在的size，线段树查询即可
yysy，我把线段树换成树状数组从1700ms变成了400ms
Code:
#include <bits/stdc++.h>
using namespace std;
#define INF 0x3f3f3f3f
#define N 100010
#define ll long long
#define reg register int
//#define int long long
char buf[1<<21], *p1=buf, *p2=buf;
#define getchar() (p1==p2&&(p2=(p1=buf)+fread(buf, 1, 1<<21, stdin)), p1==p2?EOF:*p1++)
inline int read() {
int ans=0, f=1; char c=getchar();
while (!isdigit(c)) {if (c=='-') f=-f; c=getchar();}
while (isdigit(c)) {ans=(ans<<3)+(ans<<1)+(c^48); c=getchar();}
return ans*f;
}
int n, m;
int fa[N], cnt[N];
bool vis[N];
inline int find(int p) {return fa[p]==p?p:fa[p]=find(fa[p]);}
namespace force{
void solve() {
for (int i=1; i<=n; ++i) fa[i]=i, cnt[i]=1, vis[i]=1;
ll ans;
for (int i=1,x,y,c,f1,f2; i<=m; ++i) {
if (read()&1) {
x=read(); y=read();
f1=find(x), f2=find(y);
if (f1!=f2) {
cnt[f1]+=cnt[f2];
vis[f2]=0;
fa[f2]=f1;
}
}
else {
ans=0;
c=read();
for (int i=1; i<=n; ++i) if (vis[i])
for (int j=i+1; j<=n; ++j) if (vis[j])
if (abs(cnt[i]-cnt[j])>=c) ++ans;
printf("%lld\n", ans);
}
}
exit(0);
}
}
namespace task1{
int tl[N<<2], tr[N<<2], sum[N<<2];
#define tl(p) tl[p]
#define tr(p) tr[p]
#define sum(p) sum[p]
#define pushup(p) sum(p)=sum(p<<1)+sum(p<<1|1)
void build(int p, int l, int r) {
tl(p)=l; tr(p)=r;
if (l==r) return ;
int mid=(l+r)>>1;
build(p<<1, l, mid);
build(p<<1|1, mid+1, r);
}
void upd(int p, int pos, int dat) {
if (tl(p)==tr(p)) {sum(p)+=dat; return ;}
int mid=(tl(p)+tr(p))>>1;
if (pos<=mid) upd(p<<1, pos, dat);
else upd(p<<1|1, pos, dat);
pushup(p);
}
int query(int p, int l, int r) {
if (l<=tl(p) && r>=tr(p)) return sum(p);
int mid=(tl(p)+tr(p))>>1, ans=0;
if (l<=mid) ans+=query(p<<1, l, r);
if (r>mid) ans+=query(p<<1|1, l, r);
return ans;
}
void solve() {
build(1, 1, n); upd(1, 1, n);
for (int i=1; i<=n; ++i) fa[i]=i, cnt[i]=1, vis[i]=1;
ll ans;
for (int i=1,x,y,c,f1,f2; i<=m; ++i) {
if (read()&1) {
x=read(); y=read();
f1=find(x), f2=find(y);
if (f1!=f2) {
upd(1, cnt[f1], -1); upd(1, cnt[f2], -1);
cnt[f1]+=cnt[f2];
upd(1, cnt[f1], 1);
vis[f2]=0;
fa[f2]=f1;
}
}
else {
ans=0;
c=read();
for (reg i=1,l,r; i<=n; ++i) if (vis[i]) {
if ((l=cnt[i]-c)>=1) ans+=query(1, 1, l)-(c==0);
if ((r=cnt[i]+c+(c==0))<=n) ans+=query(1, r, n);
}
printf("%lld\n", ans/2);
}
}
exit(0);
}
}
namespace task2{
int tl[N<<2], tr[N<<2], sum[N<<2], tot[N], maxn=1;
#define tl(p) tl[p]
#define tr(p) tr[p]
#define sum(p) sum[p]
#define pushup(p) sum(p)=sum(p<<1)+sum(p<<1|1)
void build(int p, int l, int r) {
tl(p)=l; tr(p)=r;
if (l==r) return ;
int mid=(l+r)>>1;
build(p<<1, l, mid);
build(p<<1|1, mid+1, r);
}
void upd(int p, int pos, int dat) {
if (tl(p)==tr(p)) {sum(p)+=dat; return ;}
int mid=(tl(p)+tr(p))>>1;
if (pos<=mid) upd(p<<1, pos, dat);
else upd(p<<1|1, pos, dat);
pushup(p);
}
int query(int p, int l, int r) {
if (l<=tl(p) && r>=tr(p)) return sum(p);
int mid=(tl(p)+tr(p))>>1, ans=0;
if (l<=mid) ans+=query(p<<1, l, r);
if (r>mid) ans+=query(p<<1|1, l, r);
return ans;
}
void solve() {
build(1, 1, n); upd(1, 1, n); tot[1]=n;
for (int i=1; i<=n; ++i) fa[i]=i, cnt[i]=1, vis[i]=1;
ll ans;
for (int i=1,x,y,c,f1,f2; i<=m; ++i) {
if (read()&1) {
x=read(); y=read();
f1=find(x), f2=find(y);
if (f1!=f2) {
upd(1, cnt[f1], -1); upd(1, cnt[f2], -1);
--tot[cnt[f1]]; --tot[cnt[f2]];
cnt[f1]+=cnt[f2];
++tot[cnt[f1]]; maxn=max(maxn, cnt[f1]);
upd(1, cnt[f1], 1);
vis[f2]=0;
fa[f2]=f1;
}
}
else {
ans=0;
c=read();
for (reg i=1,l,r; i<=maxn; ++i) if (tot[i]) {
if ((l=i-c)>=1) ans+=1ll*tot[i]*(query(1, 1, l)-(c==0));
//cout<<"l: "<<l<<' '<<tot[i]*(query(1, 1, l)-(c==0))<<endl;
if ((r=i+c+(c==0))<=n) ans+=1ll*tot[i]*query(1, r, n);
//cout<<"r: "<<r<<' '<<query(1, r, n)<<endl;
}
printf("%lld\n", ans/2);
}
}
exit(0);
}
}
namespace task{
unordered_set<int> s;
unordered_set<int>::iterator sta[N];
int sum[N], tot[N], top;
inline void upd(int i, int dat) {for (; i<=n; i+=i&-i) sum[i]+=dat;}
inline int query(int i) {int ans=0; for (; i; i-=i&-i) ans+=sum[i]; return ans;}
void solve() {
upd(1, n); tot[1]=n; s.insert(1);
for (int i=1; i<=n; ++i) fa[i]=i, cnt[i]=1, vis[i]=1;
ll ans;
for (int i=1,x,y,c,f1,f2; i<=m; ++i) {
if (read()&1) {
x=read(); y=read();
f1=find(x), f2=find(y);
if (cnt[f1]<cnt[f2]) swap(f1, f2);
if (f1!=f2) {
upd(cnt[f1], -1); upd(cnt[f2], -1);
--tot[cnt[f1]]; --tot[cnt[f2]];
cnt[f1]+=cnt[f2];
if (++tot[cnt[f1]]==1) s.insert(cnt[f1]);
upd(cnt[f1], 1);
vis[f2]=0;
fa[f2]=f1;
}
}
else {
ans=0;
c=read();
int l, r;
for (unordered_set<int>::iterator it=s.begin(); it!=s.end(); ++it) {
//cout<<"*it: "<<*it<<endl;
if (!tot[*it]) {sta[++top]=it; continue;}
if ((l=*it-c)>=1) ans+=1ll*tot[*it]*(query(l)-(c==0));
//cout<<"l: "<<l<<' '<<tot[i]*(query(1, 1, l)-(c==0))<<endl;
if ((r=*it+c+(c==0))<=n) ans+=1ll*tot[*it]*(query(n)-query(r-1));
//cout<<"r: "<<r<<' '<<query(1, r, n)<<endl;
}
while (top) s.erase(sta[top--]);
printf("%lld\n", ans/2);
}
}
exit(0);
}
}
signed main()
{
n=read(); m=read();
//if (n<=100) force::solve();
//else if (n<=1000) task1::solve();
//else task2::solve();
task::solve();
return 0;
}
