• P4234-最小差值生成树【LCT】


    正题

    题目链接:https://www.luogu.com.cn/problem/P4234


    题目大意

    给出(n)个点(m)条边的一张图。求一棵生成树使得最大边权减去最小边权最小。

    (1leq nleq 5 imes 10^4,1leq mleq 2 imes 10^5)


    解题思路

    按照边权排序,然后像魔法森林一样用(LCT)维护最小生成树就好了。

    没啥别的,练练手而已。时间复杂度(O(nlog n))


    code

    #include<cstdio>
    #include<cstring>
    #include<algorithm>
    #include<stack>
    using namespace std;
    const int N=3e5+10;
    struct node{
    	int x,y,w;
    }e[N];
    int n,m,p[N],fa[N];bool v[N];
    struct LCT{
    	int fa[N],t[N][2];
    	bool r[N];stack<int> s;
    	bool Nroot(int x)
    	{return fa[x]&&(t[fa[x]][0]==x||t[fa[x]][1]==x);}
    	bool Direct(int x)
    	{return t[fa[x]][1]==x;}
    	void PushUp(int x)
    	{p[x]=min(min(p[t[x][0]],p[t[x][1]]),x);return;}
    	void Rev(int x)
    	{r[x]^=1;swap(t[x][0],t[x][1]);return;}
    	void PushDown(int x)
    	{if(r[x])Rev(t[x][0]),Rev(t[x][1]),r[x]=0;return;}
    	void Rotate(int x){
    		int y=fa[x],z=fa[y];
    		int xs=Direct(x),ys=Direct(y);
    		int w=t[x][xs^1];
    		t[y][xs]=w;t[x][xs^1]=y;
    		if(Nroot(y))t[z][ys]=x;
    		if(w)fa[w]=y;fa[y]=x;fa[x]=z;
    		PushUp(y);PushUp(x);return;
    	}
    	void Splay(int x){
    		int y=x;s.push(x);
    		while(Nroot(y))y=fa[y],s.push(y);
    		while(!s.empty())PushDown(s.top()),s.pop();
    		while(Nroot(x)){
    			int y=fa[x];
    			if(!Nroot(y))Rotate(x);
    			else if(Direct(x)==Direct(y))
    				Rotate(y),Rotate(x);
    			else Rotate(x),Rotate(x);
    		}
    		return;
    	}
    	void Access(int x){
    		for(int y=0;x;y=x,x=fa[x])
    			Splay(x),t[x][1]=y,PushUp(x);
    		return;
    	}
    	void MakeRoot(int x)
    	{Access(x);Splay(x);Rev(x);return;}
    	int Split(int x,int y)
    	{MakeRoot(x);Access(y);Splay(y);return p[y];}
    	void Link(int x,int y)
    	{MakeRoot(x);fa[x]=y;Access(x);return;}
    	void Cut(int x,int y)
    	{MakeRoot(x);Access(y);Splay(y);fa[t[y][0]]=0;t[y][0]=0;PushUp(y);return;}
    }T;
    int find(int x)
    {return (fa[x]==x)?x:(fa[x]=find(fa[x]));}
    bool cmp(node x,node y)
    {return x.w<y.w;}
    int main()
    {
    	scanf("%d%d",&n,&m);
    	for(int i=1;i<=m;i++)
    		scanf("%d%d%d",&e[i].x,&e[i].y,&e[i].w);
    	sort(e+1,e+1+m,cmp);
    	memset(p,0x3f,sizeof(p)); 
    	for(int i=1;i<=n+m;i++)fa[i]=p[i]=i;
    	int k=n,z=0,ans=1e5;
    	for(int i=1;i<=m;i++){
    		int x=e[i].x,y=e[i].y;
    		if(x==y)continue;
    		int fx=find(x),fy=find(y);
    		if(fx==fy){
    			int num=T.Split(x+m,y+m);
    			T.Cut(e[num].x+m,num);
    			T.Cut(num,e[num].y+m);
    			v[num]=0;
    		}
    		else fa[fx]=fy,k--;
    		T.Link(x+m,i);T.Link(i,y+m);
    		v[i]=1;while(!v[z])z++;
    		if(k==1)ans=min(ans,e[i].w-e[z].w);
    	}
    	printf("%d
    ",ans);
    	return 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/QuantAsk/p/14630469.html
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