题解 CF827D Best Edge Weight
Description
You are given a connected weighted graph with n vertices and m edges. The graph doesn't contain loops nor multiple edges. Consider some edge with id i. Let's determine for this edge the maximum integer weight we can give to it so that it is contained in all minimum spanning trees of the graph if we don't change the other weights.
You are to determine this maximum weight described above for each edge. You should calculate the answer for each edge independently, it means there can't be two edges with changed weights at the same time.
If an edge is contained in every minimum spanning tree with any weight, print -1 as the answer.
\(2\leq n\leq2*10^5,n-1\leq m\leq2*10^5,c\leq 10^9\)
Solution
先求出原图的任意一棵最小生成树 , 然后把所有边分成树边和非树边进行讨论 .
对于非树边
要想出现在最小生成树上 , 就必须比当前树上连接该非树边连接的两个点的路径上最大的边的权值小 , 才能保证取代这条边 .故答案即为最小生成树上该非树边连接的两个点的路径上最大权值 \(-1\) .
对于树边 , 它必须比所有想要取代它的非树边权值小 , 才能不被取代 .故答案为所有连接的两个点的路径经过这条边的非树边的权值的最小值 \(-1\) .
用树剖跳链来维护 . 复杂度 \(O(n\log^2n)\)
Code
#include<iostream>
#include<cstdio>
#include<algorithm>
using namespace std;
int read()
{
int ret=0;char c=getchar();
while(c>'9'||c<'0')c=getchar();
while(c>='0'&&c<='9')ret=(ret<<3)+(ret<<1)+(c^48),c=getchar();
return ret;
}
const int maxn=2e5+5;
const int inf=1e9+5;
int n,m;
struct edge
{
int u,v,w;int num;
const bool operator<(const edge &a)const{return w<a.w;}
}e[maxn];
int ans[maxn];
struct segment1
{
#define L(u) u<<1
#define R(u) u<<1|1
struct node{int mx,tag;}t[maxn<<3];
void update(int u){t[u].mx=max(t[L(u)].mx,t[R(u)].mx);}
void pushdown(int u)
{
if(t[u].tag)
{
t[L(u)].mx=max(t[L(u)].mx,t[u].tag);
t[L(u)].tag=max(t[L(u)].tag,t[u].tag);
t[R(u)].mx=max(t[R(u)].mx,t[u].tag);
t[R(u)].tag=max(t[R(u)].tag,t[u].tag);
t[u].tag=0;
}
}
void change(int u,int l,int r,int nl,int nr,int x)
{
if(l>=nl&&r<=nr)
{
t[u].mx=max(t[u].mx,x);
t[u].tag=max(t[u].tag,x);
return;
}
pushdown(u);
int mid=(l+r)>>1;
if(mid>=nl)change(L(u),l,mid,nl,nr,x);
if(mid+1<=nr)change(R(u),mid+1,r,nl,nr,x);
update(u);
}
int query(int u,int l,int r,int nl,int nr)
{
if(l>=nl&&r<=nr)return t[u].mx;
pushdown(u);
int mid=(l+r)>>1,ret=0;
if(mid>=nl)ret=max(ret,query(L(u),l,mid,nl,nr));
if(mid+1<=nr)ret=max(ret,query(R(u),mid+1,r,nl,nr));
return ret;
}
}tr1;
struct segment2
{
#define L(u) u<<1
#define R(u) u<<1|1
struct node{int mi,tag;}t[maxn<<3];
void update(int u){t[u].mi=min(t[L(u)].mi,t[R(u)].mi);}
void pushdown(int u)
{
if(t[u].tag)
{
t[L(u)].mi=min(t[L(u)].mi,t[u].tag);
if(t[L(u)].tag)t[L(u)].tag=min(t[L(u)].tag,t[u].tag);
else t[L(u)].tag=t[u].tag;
t[R(u)].mi=min(t[R(u)].mi,t[u].tag);
if(t[R(u)].tag)t[R(u)].tag=min(t[R(u)].tag,t[u].tag);
else t[R(u)].tag=t[u].tag;
t[u].tag=0;
}
}
void build(int u,int l,int r)
{
t[u].mi=inf;
if(l==r)return;
int mid=(l+r)>>1;
build(L(u),l,mid);build(R(u),mid+1,r);
}
void change(int u,int l,int r,int nl,int nr,int x)
{
if(l>=nl&&r<=nr)
{
t[u].mi=min(t[u].mi,x);
if(!t[u].tag)t[u].tag=x;
else t[u].tag=min(t[u].tag,x);
return;
}
pushdown(u);
int mid=(l+r)>>1;
if(mid>=nl)change(L(u),l,mid,nl,nr,x);
if(mid+1<=nr)change(R(u),mid+1,r,nl,nr,x);
update(u);
}
int query(int u,int l,int r,int nl,int nr)
{
if(l>=nl&&r<=nr)return t[u].mi;
pushdown(u);
int mid=(l+r)>>1,ret=inf;
if(mid>=nl)ret=min(ret,query(L(u),l,mid,nl,nr));
if(mid+1<=nr)ret=min(ret,query(R(u),mid+1,r,nl,nr));
return ret;
}
}tr2;
struct mst
{
int pv[maxn],pnum[maxn];
int head[maxn],ver[maxn<<1],nxt[maxn<<1],val[maxn<<1],num[maxn<<1],tot;
void add(int x,int y,int z,int ord){ver[++tot]=y;val[tot]=z;num[tot]=ord;nxt[tot]=head[x];head[x]=tot;}
void link(int x,int y,int z,int ord){add(x,y,z,ord);add(y,x,z,ord);}
int siz[maxn],dfn[maxn],tim,top[maxn],dep[maxn],fa[maxn],son[maxn];
void getson(int u,int la)
{
siz[u]=1;fa[u]=la;dep[u]=dep[la]+1;
for(int i=head[u];i;i=nxt[i])
{
if(ver[i]==la)continue;
pv[ver[i]]=val[i];pnum[ver[i]]=num[i];
getson(ver[i],u);
siz[u]+=siz[ver[i]];
if(siz[ver[i]]>siz[son[u]])son[u]=ver[i];
}
}
void getroad(int u,int la,int ance)
{
dfn[u]=++tim;top[u]=ance;
tr1.change(1,1,n,dfn[u],dfn[u],pv[u]);
if(son[u])getroad(son[u],u,ance);
for(int i=head[u];i;i=nxt[i])
{
if(ver[i]==la||ver[i]==son[u])continue;
getroad(ver[i],u,ver[i]);
}
}
int lca(int x,int y)
{
while(top[x]!=top[y])
{
if(dep[top[x]]<dep[top[y]])swap(x,y);
x=fa[top[x]];
}
return dep[x]<dep[y]?x:y;
}
int query(int x,int y,int v)
{
int lf=lca(x,y);
int ret=0;
while(top[x]!=top[lf])
{
ret=max(ret,tr1.query(1,1,n,dfn[top[x]],dfn[x]));
tr2.change(1,1,n,dfn[top[x]],dfn[x],v);
x=fa[top[x]];
}
if(x!=lf)
{
ret=max(ret,tr1.query(1,1,n,dfn[lf]+1,dfn[x]));
tr2.change(1,1,n,dfn[lf]+1,dfn[x],v);
}
while(top[y]!=top[lf])
{
ret=max(ret,tr1.query(1,1,n,dfn[top[y]],dfn[y]));
tr2.change(1,1,n,dfn[top[y]],dfn[y],v);
y=fa[top[y]];
}
if(y!=lf)
{
ret=max(ret,tr1.query(1,1,n,dfn[lf]+1,dfn[y]));
tr2.change(1,1,n,dfn[lf]+1,dfn[y],v);
}
return ret;
}
}v;
struct dsu
{
int fa[maxn];
void prework(){for(int i=1;i<=n;i++)fa[i]=i;}
int get(int x){return x==fa[x]?x:fa[x]=get(fa[x]);}
void merge(int x,int y){fa[get(x)]=get(y);}
bool check(int x,int y){return get(x)==get(y);}
}S;
bool chose[maxn];
int main()
{
n=read();m=read();
for(int i=1;i<=m;i++){e[i].u=read();e[i].v=read();e[i].w=read();e[i].num=i;}
sort(e+1,e+m+1);
S.prework();
for(int i=1;i<=m;i++)
{
if(!S.check(e[i].u,e[i].v))
{
chose[i]=1;
v.link(e[i].u,e[i].v,e[i].w,e[i].num);
S.merge(e[i].u,e[i].v);
}
}
tr2.build(1,1,n);
v.getson(1,1);
v.getroad(1,1,1);
for(int i=1;i<=m;i++)
{
if(chose[i])continue;
ans[e[i].num]=v.query(e[i].u,e[i].v,e[i].w)-1;
}
for(int i=2;i<=n;i++)ans[v.pnum[i]]=tr2.query(1,1,n,v.dfn[i],v.dfn[i])-1;
for(int i=1;i<=m;i++)
{
if(ans[i]==inf-1)printf("-1 ");
else printf("%d ",ans[i]);
}
return 0;
}
