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  • 【刷题】BZOJ 2816 [ZJOI2012]网络

    Description

    http://www.lydsy.com/JudgeOnline/upload/zjoi2012.pdf

    Solution

    维护树上联通块的信息,支持动态加边删边
    LCT
    总共只有10种颜色,直接建10个LCT,每个LCT维护一种颜色
    LCT还是差不多
    只是第二个操作比较麻烦,得一个一个颜色地去试

    #include<bits/stdc++.h>
    #define ll long long
    #define db double
    #define ld long double
    #define lc(x) ch[(x)][0]
    #define rc(x) ch[(x)][1]
    const int MAXN=10000+10,MAXC=15;
    int n,m,c,k,Val[MAXN],d[MAXC][MAXN];
    template<typename T> inline void chkmin(T &x,T y){x=(y<x?y:x);}
    template<typename T> inline void chkmax(T &x,T y){x=(y>x?y:x);}
    struct LCT{
    	int ch[MAXN][2],fa[MAXN],rev[MAXN],Mx[MAXN];
    	inline void init()
    	{
    		memset(ch,0,sizeof(ch));
    		memset(fa,0,sizeof(fa));
    		memset(rev,0,sizeof(rev));
    		memset(Mx,0,sizeof(Mx));
    	}
    	inline bool nroot(int x)
    	{
    		return lc(fa[x])==x||rc(fa[x])==x;
    	}
    	inline void reverse(int x)
    	{
    		std::swap(lc(x),rc(x));
    		rev[x]^=1;
    	}
    	inline void pushup(int x)
    	{
    		Mx[x]=Val[x];
    		chkmax(Mx[x],Mx[lc(x)]);
    		chkmax(Mx[x],Mx[rc(x)]);
    	}
    	inline void pushdown(int x)
    	{
    		if(rev[x])
    		{
    			if(lc(x))reverse(lc(x));
    			if(rc(x))reverse(rc(x));
    			rev[x]=0;
    		}
    	}
    	inline void rotate(int x)
    	{
    		int f=fa[x],p=fa[f],c=(rc(f)==x);
    		if(nroot(f))ch[p][rc(p)==f]=x;
    		fa[ch[f][c]=ch[x][c^1]]=f;
    		fa[ch[x][c^1]=f]=x;
    		fa[x]=p;
    		pushup(f);
    		pushup(x);
    	}
    	inline void splay(int x)
    	{
    		std::stack<int> s;
    		s.push(x);
    		for(register int i=x;nroot(i);i=fa[i])s.push(fa[i]);
    		while(!s.empty())pushdown(s.top()),s.pop();
    		for(register int y=fa[x];nroot(x);rotate(x),y=fa[x])
    			if(nroot(y))rotate((lc(y)==x)==(lc(fa[y])==y)?y:x);
    		pushup(x);
    	}
    	inline void access(int x)
    	{
    		for(register int y=0;x;x=fa[y=x])splay(x),rc(x)=y,pushup(x);
    	}
    	inline void makeroot(int x)
    	{
    		access(x);splay(x);reverse(x);
    	}
    	inline int findroot(int x)
    	{
    		access(x);splay(x);
    		while(lc(x))pushdown(x),x=lc(x);
    		splay(x);
    		return x;
    	}
    	inline void split(int x,int y)
    	{
    		makeroot(x);access(y);splay(y);
    	}
    	inline void link(int x,int y)
    	{
    		makeroot(x);
    		if(findroot(y)!=x)fa[x]=y;
    	}
    	inline void cut(int x,int y)
    	{
    		makeroot(x);
    		if(findroot(y)==x&&fa[y]==x&&!lc(y))rc(x)=fa[y]=0,pushup(x);
    	}
    };
    LCT T[MAXC];
    template<typename T> inline void read(T &x)
    {
    	T data=0,w=1;
    	char ch=0;
    	while(ch!='-'&&(ch<'0'||ch>'9'))ch=getchar();
    	if(ch=='-')w=-1,ch=getchar();
    	while(ch>='0'&&ch<='9')data=((T)data<<3)+((T)data<<1)+(ch^'0'),ch=getchar();
    	x=data*w;
    }
    template<typename T> inline void write(T x,char c='')
    {
    	if(x<0)putchar('-'),x=-x;
    	if(x>9)write(x/10);
    	putchar(x%10+'0');
    	if(c!='')putchar(c);
    }
    template<typename T> inline T min(T x,T y){return x<y?x:y;}
    template<typename T> inline T max(T x,T y){return x>y?x:y;}
    int main()
    {
    	read(n);read(m);read(c);read(k);
    	for(register int i=1;i<=c;++i)T[i].init();
    	for(register int i=1;i<=n;++i)read(Val[i]);
    	for(register int i=1;i<=m;++i)
    	{
    		int u,v,w;
    		read(u);read(v);read(w);
    		w++;
    		d[w][u]++;d[w][v]++;
    		T[w].link(u,v);
    	}
    	while(k--)
    	{
    		int opt;
    		read(opt);
    		if(opt==0)
    		{
    			int x,y;
    			read(x);read(y);
    			Val[x]=y;
    			for(register int i=1;i<=c;++i)T[i].access(x),T[i].splay(x),T[i].pushup(x);
    		}
    		if(opt==1)
    		{
    			int u,v,w,dn=1;
    			read(u);read(v);read(w);
    			w++;
    			for(register int i=1;i<=c;++i)
    				if(T[i].findroot(u)==T[i].findroot(v))
    				{
    					T[i].makeroot(u),T[i].access(v),T[i].splay(v);
    					if(T[i].lc(v)!=u||T[i].rc(u))continue;
    					dn=0;
    					if(i==w)puts("Success.");
    					else if(d[w][u]>=2||d[w][v]>=2)puts("Error 1.");
    					else if(T[w].findroot(u)==T[w].findroot(v))puts("Error 2.");
    					else
    					{
    						T[i].cut(u,v);T[w].link(u,v);
    						d[i][u]--;d[i][v]--;
    						d[w][u]++;d[w][v]++;
    						puts("Success.");
    					}
    					break;
    				}
    			if(dn)puts("No such edge.");
    		}
    		if(opt==2)
    		{
    			int c,u,v;
    			read(c);read(u);read(v);
    			c++;
    			if(T[c].findroot(u)!=T[c].findroot(v))puts("-1");
    			else T[c].split(u,v),write(T[c].Mx[v],'
    ');
    		}
    	}
    	return 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/hongyj/p/8687445.html
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