• BZOJ 2733 & splay的合并


    题意:

      带权联通块,添边与查询联通块中第k大.

    SOL:

      splay合并+并查集.

      我以为splay可以用奇技淫巧来简单合并...调了一下午终于幡然醒悟...于是就只好一个一个慢慢插...什么启发式合并...说的那么高级其实那么丝帛.

      实现得非常渣...中序遍历3060ms,换成队列并没有快起来...难道是我的splay本身就不对?...

      %%%hzw的线段树...发现其实线段树合并更简单更快...被splay蒙蔽了眼睛...思路不够开阔...唉...

    Code:

      

    /*==========================================================================
    # Last modified: 2016-03-20 10:23
    # Filename: 2733.cpp
    # Description: 
    ==========================================================================*/
    #define me AcrossTheSky 
    #include <cstdio> 
    #include <cmath> 
    #include <ctime> 
    #include <string> 
    #include <cstring> 
    #include <cstdlib> 
    #include <iostream> 
    #include <algorithm> 
      
    #include <set> 
    #include <map> 
    #include <stack> 
    #include <queue> 
    #include <vector> 
     
    #define lowbit(x) (x)&(-x) 
    #define FOR(i,a,b) for((i)=(a);(i)<=(b);(i)++) 
    #define FORP(i,a,b) for(int i=(a);i<=(b);i++) 
    #define FORM(i,a,b) for(int i=(a);i>=(b);i--) 
    #define ls(a,b) (((a)+(b)) << 1) 
    #define rs(a,b) (((a)+(b)) >> 1) 
    #define getlc(a) ch[(a)][0] 
    #define getrc(a) ch[(a)][1] 
     
    #define maxn 400000 
    #define maxm 100000 
    #define pi 3.1415926535898 
    #define _e 2.718281828459 
    #define INF 1070000000 
    using namespace std; 
    typedef long long ll; 
    typedef unsigned long long ull; 
     
    template<class T> inline 
    void read(T& num) { 
        bool start=false,neg=false; 
        char c; 
        num=0; 
        while((c=getchar())!=EOF) { 
            if(c=='-') start=neg=true; 
            else if(c>='0' && c<='9') { 
                start=true; 
                num=num*10+c-'0'; 
            } else if(start) break; 
        } 
        if(neg) num=-num; 
    } 
    /*==================split line==================*/ 
    int ch[maxn][2],fa[maxn],father[maxn],sz[maxn],key[maxn];
    void pushup(int x){ sz[x]=sz[ch[x][0]]+sz[ch[x][1]]+1; }
    int kth(int x,int k){
    	while (x){
    		if (k==sz[ch[x][0]]+1) return x;
    		if (k>sz[ch[x][0]]+1) k-=sz[ch[x][0]],k--,x=ch[x][1];
    		else x=ch[x][0];
    	}
    }
    int find(int x){
    	return father[x]==x?x:find(father[x]);
    }
    void rotate(int x){
    	int p=fa[x],q=fa[p],d=ch[p][1]==x;
    	fa[ch[p][d]=ch[x][d^1]]=p; pushup(p);
    	fa[ch[x][d^1]=p]=x;pushup(x);
    	fa[x]=q;
    	if (q){
    		if (ch[q][1]==p) ch[q][1]=x;
    		else ch[q][0]=x;
    	}
    }
    void splay(int x,int aim){
    	for (int y;(y=fa[x])!=aim;rotate(x))
    		if (fa[y]!=aim)
    			rotate((getlc(y)==x)==(getlc(fa[y])==y)?y:x);
    }
    void insert(int x,int root){
    	int node=root;
    	while (node){
    		sz[node]++; 
    		int &d=ch[node][key[node]<key[x]];
    		if (d==0) {
    			d=x; fa[x]=node; ch[x][0]=ch[x][1]=0; sz[x]=1; node=x; break;
    		}
    		node=d;
    	}
    	//splay(node,0);
    }
    int q[maxn];
    void join(int x,int root){
    	/*int lc=ch[x][0],rc=ch[x][1];
    	ch[x][0]=ch[x][1]=0;
    	if (lc) {
    		join(lc,root);
    	}
    	 //ch[x][1]=0; fa[rc]=0;
    	insert(x,root);
    	if (rc) join(rc,root);*/
    	int head=1,tail=1;
    	q[head]=x;
    	while (head<=tail){
    		x=q[head]; head++;
    		int lc=ch[x][0],rc=ch[x][1];
    		ch[x][0]=ch[x][1]=0;
    		insert(x,root);
    		if (lc) q[++tail]=lc; 
    		if (rc) q[++tail]=rc;
    	}
    } 
    void Union(int x,int y){
    	int u=find(x),v=find(y);
    	if (u==v) return;
    	else {
    		father[u]=v;
    		splay(u,0); splay(v,0);
    		if (sz[u]<sz[v]) join(u,v);
    			else join(v,u);/*father[u]=v;
    		int a=kth(u,1),b=kth(v,sz[v]);
    		splay(a,0); splay(b,0);
    		if (key[a]<key[b]) {ch[a][1]=b; fa[b]=a; pushup(a);}
    			else {ch[b][1]=a; fa[a]=b; pushup(b);}
    		*/
    
    	}
    }
    void query(int x,int k){
    	splay(x,0);
    	if (k>sz[x]) printf("-1
    ");
    	else printf("%d
    ",kth(x,k));
    }
    int main(){
    	int n,m; read(n); read(m);
    	memset(father,0,sizeof(father));
    	FORP(i,1,n) { read(key[i]); father[i]=i; sz[i]=1;}
    	FORP(i,1,m){
    		int u,v; read(u); read(v); Union(u,v);
    	}
    	int q; read(q);
    	FORP(i,1,q){
    		char ch[10]; scanf("%s",ch);
    		if (ch[0]=='Q') {
    			int x,k; read(x); read(k);
    			query(x,k);
    		}
    		else { int u,v; read(u); read(v); Union(u,v);}
    	}
    }
    
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  • 原文地址:https://www.cnblogs.com/YCuangWhen/p/5300523.html
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