正题

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

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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。

$1\le n\le 5\times 10^4,1\le m\le 2\times 10^5$

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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$O(n\log n)$

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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\n",ans);return 0;
}