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。
\(1\leq n\leq 5\times 10^4,1\leq m\leq 2\times 10^5\)
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\(O(n\log n)\)
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\n",ans);
return 0;
}