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树链剖分树套树

这里的树套树使用树状数组动态开点权值线段树实现,要先离散化

我的方法算是有点笨但是好写的方法

#include<bits/stdc++.h>
#pragma GCC optimize(3)
#define il __inline__ __attribute__ ((always_inline))
#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)
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;
const int N=80011;
int n,q,cnt,nn,a[N],head[N],c[N<<1];
struct node{int v,id,typ;bool operator<(node t){return v<t.v;}}b[N<<1];
struct que{int k,x,y;}Q[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;}
int siz[N],top[N],f[N],d[N],id[N],sz;
void dfs(int x){
    siz[x]=1;
    fl(i,x)if(f[x]^to){
        d[to]=d[x]+1;
        f[to]=x;
        dfs(to);
        siz[x]+=siz[to];
    }
}
void bt(int x,int tp){
    top[x]=tp;id[x]=++sz;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);
}
int RT[N],s[N*400],ls[N*400],rs[N*400];
void upd(int x,int v,int l,int r,int &rt){
    if(!rt)rt=++sz;
    s[rt]+=v;
    if(l==r)return;
    int m=l+r>>1;
    if(x<=m)upd(x,v,l,m,ls[rt]);
    else upd(x,v,m+1,r,rs[rt]);
}
int tl[400],tr[400],cl,cr;
il void get(int l,int r){
    for(int i=l-1;i;i^=i&-i)tl[++cl]=RT[i];
    for(int i=r;i;i^=i&-i)tr[++cr]=RT[i];
}
il void ask(int x,int y,int k){
    cl=cr=0;
    int t=0;
    while(top[x]^top[y]){
        if(d[top[x]]<d[top[y]])x^=y,y^=x,x^=y;
        t+=id[x]-id[top[x]]+1;get(id[top[x]],id[x]),x=f[top[x]];
    }
    if(id[x]>id[y])x^=y,y^=x,x^=y;
    get(id[x],id[y]);
    t+=id[y]-id[x]+1;
    if(t<k){out("invalid request!\n");return;}
    int l=1,r=nn;
    while(l<r){
        int m=l+r>>1,sum=0;
        fur(i,1,cl)sum-=s[rs[tl[i]]];
        fur(i,1,cr)sum+=s[rs[tr[i]]];
        if(k<=sum){
            fur(i,1,cl)tl[i]=rs[tl[i]];
            fur(i,1,cr)tr[i]=rs[tr[i]];
            l=m+1;
        }
        else{
            fur(i,1,cl)tl[i]=ls[tl[i]];
            fur(i,1,cr)tr[i]=ls[tr[i]];
            k-=sum,r=m;
        }
    }
    out(c[l],ln);
}
int main(){
    in(n,q);
    int k,x,y,t=0;
    fur(i,1,n)in(a[i]),b[++t]={a[i],i,0};
    fur(i,2,n)in(x,y),add(x,y),add(y,x);
    fur(i,1,q){
        in(k,x,y);
        if(!k)b[++t]={y,i,1};
        Q[i]={k,x,y};
    }
    sort(b+1,b+t+1);b[0].v=-(1<<30);
    fur(i,1,t)c[(b[i].typ?Q[b[i].id].y:a[b[i].id])=nn+=b[i].v!=b[i-1].v]=b[i].v;
    dfs(1);bt(1,1);
    fur(i,1,n)for(int j=id[i];j<=n;j+=j&-j)upd(a[i],1,1,nn,RT[j]);
    fur(i,1,q){
        k=Q[i].k,x=Q[i].x,y=Q[i].y;
        if(!k){
            for(int j=id[x];j<=n;j+=j&-j)upd(a[x],-1,1,nn,RT[j]);
            a[x]=y;
            for(int j=id[x];j<=n;j+=j&-j)upd(a[x],1,1,nn,RT[j]);
        }
        else ask(x,y,k);
    }
    flush();
}
LG 4175 [CTSC2008]网络管理
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