Codeforces 1172E Nauuo and ODT [LCT]
ZROI那题是这题删掉修改的弱化版……ZROI还我培训费/px
思路
按照套路,我们考虑每种颜色的贡献,然后发现不包含某种颜色的路径条数更容易数,就是删掉该颜色的点后每个连通块大小的平方和。
由于每种颜色影响到的点之和是\(O(n)\)的,所以我们每种颜色分开考虑,就变成了只有两种颜色的情况。
把这种颜色视为白色,其他颜色视为黑色,那么每个黑色点往父亲连边(1往一个虚拟点连边),就可以在白点处统计贡献。
所以每次改变颜色就只是一个link或cut,用LCT维护。
代码
#include<bits/stdc++.h>
clock_t t=clock();
namespace my_std{
using namespace std;
#define pii pair<int,int>
#define fir first
#define sec second
#define MP make_pair
#define rep(i,x,y) for (int i=(x);i<=(y);i++)
#define drep(i,x,y) for (int i=(x);i>=(y);i--)
#define go(x) for (int i=head[x];i;i=edge[i].nxt)
#define templ template<typename T>
#define sz 404004
typedef long long ll;
typedef double db;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
templ inline T rnd(T l,T r) {return uniform_int_distribution<T>(l,r)(rng);}
templ inline bool chkmax(T &x,T y){return x<y?x=y,1:0;}
templ inline bool chkmin(T &x,T y){return x>y?x=y,1:0;}
templ inline void read(T& t)
{
t=0;char f=0,ch=getchar();double d=0.1;
while(ch>'9'||ch<'0') f|=(ch=='-'),ch=getchar();
while(ch<='9'&&ch>='0') t=t*10+ch-48,ch=getchar();
if(ch=='.'){ch=getchar();while(ch<='9'&&ch>='0') t+=d*(ch^48),d*=0.1,ch=getchar();}
t=(f?-t:t);
}
template<typename T,typename... Args>inline void read(T& t,Args&... args){read(t); read(args...);}
char __sr[1<<21],__z[20];int __C=-1,__zz=0;
inline void Ot(){fwrite(__sr,1,__C+1,stdout),__C=-1;}
inline void print(register int x)
{
if(__C>1<<20)Ot();if(x<0)__sr[++__C]='-',x=-x;
while(__z[++__zz]=x%10+48,x/=10);
while(__sr[++__C]=__z[__zz],--__zz);__sr[++__C]='\n';
}
void file()
{
#ifdef NTFOrz
freopen("a.in","r",stdin);
#endif
}
inline void chktime()
{
#ifndef ONLINE_JUDGE
cout<<(clock()-t)/1000.0<<'\n';
#endif
}
#ifdef mod
ll ksm(ll x,int y){ll ret=1;for (;y;y>>=1,x=x*x%mod) if (y&1) ret=ret*x%mod;return ret;}
ll inv(ll x){return ksm(x,mod-2);}
#else
ll ksm(ll x,int y){ll ret=1;for (;y;y>>=1,x=x*x) if (y&1) ret=ret*x;return ret;}
#endif
// inline ll mul(ll a,ll b){ll d=(ll)(a*(double)b/mod+0.5);ll ret=a*b-d*mod;if (ret<0) ret+=mod;return ret;}
}
using namespace my_std;
int n,m;
int col[sz];
struct hh{int t,nxt;}edge[sz<<1];
int head[sz],ecnt;
void make_edge(int f,int t)
{
edge[++ecnt]=(hh){t,head[f]};
head[f]=ecnt;
edge[++ecnt]=(hh){f,head[t]};
head[t]=ecnt;
}
int f[sz];
void dfs(int x,int fa)
{
f[x]=fa;
go(x) if (edge[i].t!=fa) dfs(edge[i].t,x);
}
#define Sq(x) (1ll*(x)*(x))
int nc[sz];
namespace LCT
{
int fa[sz],s[sz],size[sz],ch[sz][2],rt[sz];
#define S(x) (size[x]+s[x])
ll sq[sz];
#define ls ch[x][0]
#define rs ch[x][1]
int get(int x){return ch[fa[x]][1]==x;}
int nroot(int x){return ch[fa[x]][0]==x||ch[fa[x]][1]==x;}
void pushup(int x){s[x]=S(ls)+S(rs)+1;if (ls) rt[x]=rt[ls]; else rt[x]=x;}
void rotate(int x)
{
int y=fa[x],z=fa[y],k=get(x),w=ch[x][!k];
if (nroot(y)) ch[z][get(y)]=x;ch[x][!k]=y;ch[y][k]=w;
if (w) fa[w]=y;fa[y]=x;fa[x]=z;
pushup(y),pushup(x);
}
void splay(int x){for (int y;y=fa[x],nroot(x);rotate(x)) if (nroot(y)) rotate(get(x)==get(y)?y:x);}
void access(int x){for (int y=0;x;x=fa[y=x]) splay(x),size[x]+=S(rs)-S(y),sq[x]+=Sq(S(rs))-Sq(S(y)),rs=y,pushup(x);}
ll calc(int x){splay(x);return sq[x]+Sq(S(rs));}
ll link(int x)
{
nc[x]=0;
int y=f[x];
access(x),splay(x);
access(y),splay(y);
int root=rt[y];splay(root);
ll ret=-calc(root)-calc(x);
fa[x]=y;size[y]+=S(x);sq[y]+=Sq(S(x));splay(y);
ret+=calc(root);
return ret;
}
ll cut(int x)
{
nc[x]=1;
int y=f[x];
access(x),splay(y);
int root=rt[y];splay(root);
ll ret=-calc(root);
splay(y);ch[y][1]=fa[x]=0;pushup(y);
ret+=calc(x);
ret+=calc(root);
return ret;
}
}
using namespace LCT;
ll ans[sz];
vector<pii>V[sz];
int main()
{
file();
read(n,m);
rep(i,1,n) read(col[i]),V[col[i]].push_back(MP(i,0));
int x,y;
rep(i,1,n-1) read(x,y),make_edge(x,y);
dfs(1,0);
rep(i,1,m) read(x,y),V[col[x]].push_back(MP(x,i)),V[col[x]=y].push_back(MP(x,i));
f[1]=n+1;nc[n+1]=1;
rep(i,1,n) pushup(i);
rep(i,1,n) link(i);
rep(c,1,n)
{
ans[0]+=1ll*n*n;
for (auto w:V[c])
{
int x=w.fir;
if (nc[x]) ans[w.sec]+=link(x);
else ans[w.sec]+=cut(x);
}
for (auto w:V[c]) if (nc[w.fir]) link(w.fir);
}
rep(i,1,m) ans[i]+=ans[i-1];
rep(i,0,m) printf("%I64d\n",1ll*n*n*n-ans[i]);
return 0;
}