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2020-01-06 01:03:00
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先考虑没有换根的操作(设根节点为1)

修改点x的权值

被影响的只有1\rightarrow x上的点,设p_1,p_2,...,p_k1\rightarrow x上的点

s_ii子树点权和

将修改看成增加v

修改后贡献为

\sum_{i=1}^k (s_{p_i}+v)^2-\sum_{i=1}^k s_{p_i}^2= k\times v^2+2v\sum_{i=1}^ks_{p_i}

我们需要预处理出s_i^2,维护s_i

考虑换根操作:

x为根,p_1,p_2,...,p_k1\rightarrow x上的点

a_i为在1为根时v_i的点权和,b_i为在x为根时v_i的点权和

可以发现a_{i+1}+b_i=a_1=b_k(因为所有点的点权和总是相同)

ans_x表示x为根时的答案

ans_x=ans_1+\sum_{i=1}^k b_i^2-a_i^2

=ans_1+\sum_{i=1}^k (a_1-a_{i+1})^2-a_i^2

=ans_1+\sum_{i=2}^k (a_1-a_i)^2-a_i^2

=ans_1+ (k-1)a_1^2- 2a_1\sum_{i=2}^k a_i

=ans_1+ s_1[(k+1)s_1-2\sum_{i=1}^k s_{p_i}]

可以使用树链剖分+(树状数组/线段树)或LCT维护

#include<bits/stdc++.h>
namespace ZDY{
    #pragma GCC optimize(3)
    #define il __inline__ __attribute__ ((always_inline))
    #define register
    #define ll long long
    #define ull unsigned long long
    #define db double
    #define sht short
    #define MB template <class T>il
    #define Fur(i,x,y) for(int i(x);i<=y;++i)
    #define Fdr(i,x,y) for(int i(x);i>=y;--i)
    #define fl(i,x) for(int i(head[x]),to;to=e[i].to,i;i=e[i].nxt)
    #define clr(x,y) memset(x,y,sizeof(x))
    #define cpy(x,y) memcpy(x,y,sizeof(x))
    #define fin(s) freopen(s".in","r",stdin)
    #define fout(s) freopen(s".out","w",stdout)
    #define fcin ios::sync_with_stdio(false)
    #define l2(n) ((int)(log2(n)))
    #define inf 2122219134
    MB T ABS(T x){return x>0?x:-x;}
    MB T MAX(T x,T y){return x>y?x:y;}
    MB T MIN(T x,T y){return x<y?x:y;}
    MB T GCD(T x,T y){return y?GCD(y,x%y):x;}
    MB void SWAP(T &x,T &y){T t=x;x=y;y=t;}
}using namespace ZDY;using namespace std;
namespace IO{const int str=1<<20;static char in_buf[str],*in_s,*in_t;bool __=0;il char gc(){return (in_s==in_t)&&(in_t=(in_s=in_buf)+fread(in_buf,1,str,stdin)),in_s==in_t?EOF:*in_s++;}il void in(string &ch){ch.clear();if(__)return;char c;while((c=gc())!=EOF&&isspace(c));if(c==EOF){__=1;return;}ch+=c;while((c=gc())!=EOF&&!isspace(c))ch+=c;if(c==EOF)__=1;}il void in(char &ch){if(__)return;char c;while((c=gc())!=EOF&&isspace(c));if(c==EOF)__=1;else ch=c;}il void in(char *ch){*ch='\0';if(__)return;char c;while((c=gc())!=EOF&&isspace(c));if(c==EOF){__=1;return;}*ch=c;ch++;while((c=gc())!=EOF&&!isspace(c))*ch=c,ch++;if(c==EOF)__=1;*ch='\0';}template<typename T>il void in(T &x){if(__)return;char c=gc();bool f=0;while(c!=EOF&&(c<'0'||c>'9'))f^=(c=='-'),c=gc();if(c==EOF){__=1;return;}x=0;while(c!=EOF&&'0'<=c&&c<='9')x=x*10+c-48,c=gc();if(c==EOF)__=1;if(f)x=-x;}template<typename T,typename ... arr>il void in(T &x,arr & ... y){in(x),in(y...);}const char ln='\n';static char out_buf[str],*out_s=out_buf,*out_t=out_buf+str;il void flush(){fwrite(out_buf,1,out_s-out_buf,stdout);out_s=out_buf;}il void pt(char c){(out_s==out_t)?(fwrite(out_s=out_buf,1,str,stdout),*out_s++=c):(*out_s++=c);}il void out(const char* s){while(*s)pt(*s++);}il void out(char* s){while(*s)pt(*s++);}il void out(char c){pt(c);}il void out(string s){for(int i=0;s[i];i++)pt(s[i]);}template<typename T>il void out(T x){if(!x){pt('0');return;}if(x<0)pt('-'),x=-x;char a[50],t=0;while(x)a[t++]=x%10,x/= 10;while(t--)pt(a[t]+'0');}template<typename T,typename ... arr>il void out(T x,arr & ... y){out(x),out(y...);}}using namespace IO;
#define N 200011
int n,cnt=0,head[N],val[N],pos[N],w[N];
struct edge{
    int to,nxt;
}e[N<<1];
il void add(int x,int y){
    e[++cnt].to=y;e[cnt].nxt=head[x];head[x]=cnt;
}
ll s[N],S[N],ans=0;
il void upd(int x,int v){
    if(x<=n)
    for(int i=x;i<=n;i+=i&-i)
        s[i]+=v,
        S[i]+=1ll*x*v;
}
il void upd(int l,int r,int v){upd(l,v);upd(r+1,-v);}
il ll ask(int x){
    ll res=0;
    for(int i=x;i;i-=i&-i)
        res+=1ll*(x+1)*s[i]-S[i];
    return res;
}
il ll ask(int l,int r){return ask(r)-ask(l-1);}
int siz[N],top[N],id[N],f[N],d[N],sz=0;
void dfs(int x){
    siz[x]=1;
    fl(i,x)if(to!=f[x]){
        f[to]=x;d[to]=d[x]+1;
        dfs(to);
        siz[x]+=siz[to];
        w[x]+=w[to];
    }
}
void bt(int x,int tp){
    top[x]=tp;pos[id[x]=++sz]=x;
    ans+=1ll*w[x]*w[x];
    upd(id[x],w[x]-w[pos[sz-1]]);
    int k=0;
    fl(i,x)if(to!=f[x]&&siz[to]>siz[k])k=to;
    if(!k)return;bt(k,tp);
    fl(i,x)if(!top[to])bt(to,to);
}
il void mdy(int x,int v){
    int k=0;
    ll res=0;
    while(x){
        k+=id[x]-id[top[x]]+1;
        res+=ask(id[top[x]],id[x]);
        upd(id[top[x]],id[x],v);
        x=f[top[x]];
    }
    ans+=1ll*v*(1ll*v*k+(res<<1));
}
il ll qry(int x){
    int k=0,v=ask(1);
    ll res=0;
    while(x){
        k+=id[x]-id[top[x]]+1;
        res+=ask(id[top[x]],id[x]);
        x=f[top[x]];
    }
    return ans+1ll*v*((k+1)*v-(res<<1));
}
int main(){
    int q,x,y,opt;
    in(n,q);
    Fur(i,1,n-1)
        in(x,y),
        add(x,y),add(y,x);
    Fur(i,1,n)in(val[i]),w[i]=val[i];
    dfs(1);bt(1,1);
    while(q--){
        in(opt,x);
        if(opt==1){
            in(y);
            y-=val[x];
            val[x]+=y;
            mdy(x,y);
        }
        else out(qry(x),ln);
    }
    flush();
}
LG 3676 小清新数据结构题
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