标签:ons using highlight empty push NPU nbsp add ret
类似贪心,用 BFS 对树进行染色,然后枚举哪些边的两个端点颜色不同.
code:
#include <bits/stdc++.h>
#define N 300006
#define setIO(s) freopen(s".in","r",stdin)
using namespace std;
vector<int>G;
queue<int>q;
int n,k,d,edges,vis[N],hd[N],to[N<<1],nex[N<<1],U[N],V[N],dis[N];
void add(int u,int v)
{
nex[++edges]=hd[u],hd[u]=edges,to[edges]=v;
}
int main()
{
int i,j;
// setIO("input");
scanf("%d%d%d",&n,&k,&d);
for(i=1;i<=k;++i)
{
int x;
scanf("%d",&x);
vis[x]=x,q.push(x);
}
for(i=1;i<n;++i)
{
int u,v;
scanf("%d%d",&u,&v), add(u,v),add(v,u);
U[i]=u, V[i]=v;
}
for(;!q.empty();)
{
int u=q.front();q.pop();
for(int i=hd[u];i;i=nex[i])
{
int v=to[i];
if(!vis[v] && dis[u]+1<=d)
{
vis[v]=vis[u];
dis[v]=dis[u]+1;
q.push(v);
}
}
}
for(i=1;i<n;++i)
{
if(vis[U[i]]!=vis[V[i]])
{
G.push_back(i);
}
}
printf("%d\n",G.size());
for(i=0;i<G.size();++i) printf("%d ",G[i]);
return 0;
}
标签:ons using highlight empty push NPU nbsp add ret
原文地址:https://www.cnblogs.com/guangheli/p/11619417.html