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  • bzoj2002:[Hnoi2010]Bounce 弹飞绵羊

    思路:动态树。把每个装置看成点,能弹到的装置连边,能弹飞的装置连到n+1号点,每个点点权设为1,这样问几次弹飞就可以转化为该点到n+1号点的路径的点权和。

    #include<cstdio>
    #include<cstring>
    #include<iostream>
    #include<algorithm>
    const int maxn=200010;
    using namespace std;
    int n,m,next[maxn];char ch;
    
    void read(int &x){
    	for (ch=getchar();!isdigit(ch);ch=getchar());
    	for (x=0;isdigit(ch);ch=getchar()) x=x*10+ch-'0';
    }
    
    struct LCT{
    	int c[maxn][2],fa[maxn],size[maxn],rev[maxn];
    	inline void update(int x){size[x]=size[c[x][0]]+size[c[x][1]]+1;}
    	inline bool isroot(int x){return (c[fa[x]][0]!=x)&&(c[fa[x]][1])!=x;}
    	inline int which(int x){return c[fa[x]][1]==x;}
    	inline void flip(int x){rev[x]^=1,swap(c[x][0],c[x][1]);}
    	inline void down(int x){if (rev[x]) flip(c[x][0]),flip(c[x][1]),rev[x]=0;}
    	void relax(int x){if (!isroot(x)) relax(fa[x]);down(x);}
    	void rotate(int x){
    		int y=fa[x],z=fa[y],nx=which(x),ny=which(y);
    		fa[c[x][!nx]]=y,c[y][nx]=c[x][!nx];
    		fa[x]=z;if (!isroot(y)) c[z][ny]=x;
    		fa[y]=x,c[x][!nx]=y;update(y);
    	}
    	void splay(int x){
    		relax(x);
    		while (!isroot(x)){
    			if (isroot(fa[x])) rotate(x);
    			else if (which(x)==which(fa[x])) rotate(fa[x]),rotate(x);
    			else rotate(x),rotate(x);
    		}
    		update(x);
    	}
    	void access(int x){for (int p=0;x;x=fa[x]) splay(x),fa[c[x][1]=p]=x,update(x),p=x;}
    	void makeroot(int x){access(x),splay(x),flip(x),update(x);}
    	void link(int a,int b){makeroot(a),fa[a]=b;}
    	void cut(int a,int b){makeroot(a),access(b),splay(b),c[b][0]=fa[a]=0;}
    	void query(int a){makeroot(n+1),access(a),splay(a),printf("%d
    ",size[c[a][0]]);}
    	void modify(int a,int k){int t=a+k>n?n+1:a+k;cut(a,next[a]),link(a,t),next[a]=t;}
    }T;
    
    int main(){
    	read(n),T.size[n+1]=1;
    	for (int i=1,a;i<=n;i++) read(a),T.fa[i]=min(i+a,n+1),T.size[i]=1,next[i]=T.fa[i];
    	read(m);
    	for (int i=1,op,a,k;i<=m;i++){
    		read(op),read(a),a++;
    		if (op==1) T.query(a);
    		else read(k),T.modify(a,k);
    	}
    	return 0;
    }


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  • 原文地址:https://www.cnblogs.com/thythy/p/5493601.html
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