Duff in the Army CodeForces - 587C

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利用vector保存信息 注意题目说了"1 ≤ a ≤ 10" 也就是说 每个vector变量里最多存10个元素就足够了 不然MLE

#include <bits/stdc++.h>
using namespace std;

struct node
{
	int v;
	int next;
};

node edge[200010];
vector <int> per[100010][20];
vector <int> pre[100010];
vector <int> ans;
int dp[100010][20];
int first[100010],deep[100010];
int n,m,num;

void addedge(int u,int v)
{
	edge[num].v=v;
	edge[num].next=first[u];
	first[u]=num++;
	return;
}

void dfs(int cur,int pre)
{
	int i,v;
	for(i=first[cur];i!=-1;i=edge[i].next)
	{
		v=edge[i].v;
		if(v!=pre)
		{
			dp[v][0]=cur;
			deep[v]=deep[cur]+1;
			dfs(v,cur); 
		}
	}
	return;
}

void init()
{
	vector <int> ::iterator iter;
	int i,j,k,t;
	for(i=1;i<=n;i++)
	{
		sort(pre[i].begin(),pre[i].end());
		pre[i].erase(unique(pre[i].begin(),pre[i].end()),pre[i].end());
	}
	for(i=1;i<=n;i++)
	{
		for(k=0;k<pre[i].size();k++)
		{
			per[i][0].push_back(pre[i][k]);
		}
		for(k=0;k<pre[dp[i][0]].size();k++)
		{
			per[i][0].push_back(pre[dp[i][0]][k]);
		}
		sort(per[i][0].begin(),per[i][0].end());
		per[i][0].erase(unique(per[i][0].begin(),per[i][0].end()),per[i][0].end());
		iter=per[i][0].begin(),t=per[i][0].size();
		for(k=0;k<min(t,10);k++) iter++;
		per[i][0].erase(iter,per[i][0].end());
	}
	for(j=1;j<=log2(n);j++)
	{
		for(i=1;i<=n;i++)
		{
			dp[i][j]=dp[dp[i][j-1]][j-1];
			for(k=0;k<per[i][j-1].size();k++)
			{
				per[i][j].push_back(per[i][j-1][k]);
			}
			for(k=0;k<per[dp[i][j-1]][j-1].size();k++)
			{
				per[i][j].push_back(per[dp[i][j-1]][j-1][k]);
			}
			sort(per[i][j].begin(),per[i][j].end());
			per[i][j].erase(unique(per[i][j].begin(),per[i][j].end()),per[i][j].end());
			iter=per[i][j].begin(),t=per[i][j].size();
			for(k=0;k<min(t,10);k++) iter++;
			per[i][j].erase(iter,per[i][j].end());
		}
	}
	return;
}

int getlca(int u,int v)
{
	int i;
	if(deep[u]<deep[v]) swap(u,v);
	for(i=log2(n);i>=0;i--)
	{
		if(deep[dp[u][i]]>=deep[v])
		{
			u=dp[u][i];
		}
	}
	if(u==v) return u;
	for(i=log2(n);i>=0;i--)
	{
		if(dp[u][i]!=dp[v][i])
		{
			u=dp[u][i],v=dp[v][i];
		}
	}
	return dp[u][0];
}

void solve(int u,int lca)
{
	int i,j;
	for(i=log2(n);i>=0;i--)
	{
		if(deep[dp[u][i]]>=deep[lca])
		{
			for(j=0;j<per[u][i].size();j++)
			{
				ans.push_back(per[u][i][j]);
			}
			u=dp[u][i];
		}
	}
	return;
}

int main()
{
	int q,i,u,v,c,t,lca;
	scanf("%d%d%d",&n,&m,&q);
	memset(first,-1,sizeof(first));
	num=0;
	for(i=1;i<=n-1;i++)
	{
		scanf("%d%d",&u,&v);
		addedge(u,v);
		addedge(v,u);
	}
	for(i=1;i<=m;i++)
	{
		scanf("%d",&c);
		if(pre[c].size()<10)
		{
			pre[c].push_back(i);
		}
	}
	
	deep[1]=1;
	dfs(1,0);
	init();
	
	while(q--)
	{
		scanf("%d%d%d",&u,&v,&c);
		lca=getlca(u,v);
		ans.clear();
		if(u==v)
		{
			t=pre[u].size();
			cout<<min(t,c)<<" ";
			for(i=0;i<min(t,c);i++)
			{
				cout<<pre[u][i]<<" ";
			}
			cout<<endl;
		}
		else
		{
			solve(u,lca);
			solve(v,lca);
			sort(ans.begin(),ans.end());
			ans.erase(unique(ans.begin(),ans.end()),ans.end());
			t=ans.size();
			cout<<min(t,c)<<" ";
			for(i=0;i<min(t,c);i++)
			{
				cout<<ans[i]<<" ";
			}
			cout<<endl;
		}
	}
		
	return 0;
}