LCT边权维护经典题
咋维护呢？边化为点，边权变点权。
本题中我们把边对关键字A进行排序，动态维护关键字B的最小生成树
加边后出现环咋办？
splay维护最大边的编号，找到最大边删除再加新边就ok辣
#include<cstdio> #include<algorithm> using namespace std; inline void Swap(int &a,int &b){a^=b^=a^=b;} inline int Min(int a,int b){return a<b?a:b;} void read(int &x){ static char c=getchar();x=0; while(c<'0'||c>'9') c=getchar(); while('0'<=c&&c<='9') x=x*10+(c^48),c=getchar(); } #define N 200005 struct edge{int f,t,A,B;}a[N]; int n,m,ans=2e9,ch[N][2],fa[N],val[N],s[N],rev[N],dad[N]; #define lc ch[x][0] #define rc ch[x][1] inline bool cmp(edge P,edge Q){return P.A<Q.A;} int find(int x){return dad[x]==x?x:dad[x]=find(dad[x]);} inline bool nrt(int x){return ch[fa[x]][0]==x||ch[fa[x]][1]==x;} void up(int x){//维护平衡树上最大点权的点的编号（就是边化为的点） s[x]=x; if(val[s[lc]]>val[s[x]]) s[x]=s[lc]; if(val[s[rc]]>val[s[x]]) s[x]=s[rc]; } void Rev(int x){Swap(lc,rc); rev[x]^=1;} void down(int x){if(rev[x]) Rev(lc),Rev(rc),rev[x]=0;} void Pre(int x){if(nrt(x))Pre(fa[x]); down(x);} void turn(int x){ int y=fa[x],z=fa[y],l=(ch[y][1]==x),r=l^1; if(nrt(y)) ch[z][ch[z][1]==y]=x; fa[ch[x][r]]=y; fa[y]=x; fa[x]=z; ch[y][l]=ch[x][r]; ch[x][r]=y; up(y); up(x); } void splay(int x){ Pre(x); for(;nrt(x);turn(x)){ int y=fa[x],z=fa[y]; if(nrt(y)) turn(((ch[z][1]==y)^(ch[y][1]==x))?x:y); } } void access(int x){for(int y=0;x;y=x,x=fa[x]) splay(x),rc=y,up(x);} void makert(int x){access(x);splay(x);Rev(x);} int findrt(int x){ access(x);splay(x);down(x); while(lc) x=lc,down(x); splay(x); return x; } void link(int x,int y){makert(x); if(findrt(y)!=x)fa[x]=y;} void cut(int x,int y){ makert(x); if(findrt(y)==x&&fa[y]==x&&!ch[y][0]) fa[y]=rc=0,up(x); } void split(int x,int y){makert(x);access(y);splay(y);} int main(){ read(n);read(m); register int i; for(i=1;i<=n+m;++i) dad[i]=i; for(i=1;i<=m;++i) read(a[i].f),read(a[i].t),read(a[i].A),read(a[i].B); sort(a+1,a+m+1,cmp); for(i=1;i<=m;++i) val[n+i]=a[i].B; for(i=1;i<=m;++i){ int r1=find(a[i].f),r2=find(a[i].t),e;//并查集维护是否连通 if(r1==r2){ split(a[i].f,a[i].t); e=s[a[i].t]; if(val[e]<=a[i].B) continue; cut(a[e-n].f,e); cut(e,a[e-n].t);//删掉最大边 }else dad[r1]=r2; link(a[i].f,i+n); link(i+n,a[i].t); if(find(1)==find(n)) split(n,1),ans=Min(ans,a[i].A+val[s[1]]); } if(ans>=2e9) puts("-1"); else printf("%d",ans); return 0; }