HDU 4912(Paths on the tree-树上取链，贪心)

Paths on the tree

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1161    Accepted Submission(s): 384

Problem Description

bobo has a tree, whose vertices are conveniently labeled by 1,2,…,n.

There are m paths on the tree. bobo would like to pick some paths while any two paths do not share common vertices.

Find the maximum number of paths bobo can pick.

 

Input

The input consists of several tests. For each tests:

The first line contains n,m (1≤n,m≤10 ⁵). Each of the following (n - 1) lines contain 2 integers a _i,b _i denoting an edge between vertices a _i and b _i (1≤a _i,b _i≤n). Each of the following m lines contain 2 integers u _i,v _i denoting a path between vertices u _i and v _i (1≤u _i,v _i≤n).

 

Output

For each tests:

A single integer, the maximum number of paths.

 

Sample Input

   
   
   
   

    
    
    
    3 2
1 2
1 3
1 2
1 3
7 3
1 2
1 3
2 4
2 5
3 6
3 7
2 3
4 5
6 7
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    1
2
   
   
   
   

 

Author

Xiaoxu Guo (ftiasch)

 

Source

2014 Multi-University Training Contest 5

 

Recommend

We have carefully selected several similar problems for you:   5426  5425  5424  5423  5422 

 

贪心，从lca由深入浅地取，

能取就取，并把lca所在子树标记为无法取

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<iostream>
#include<functional>
#include<cmath>
#include<algorithm>
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define ForkD(i,k,n) for(int i=n;i>=k;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=pre[x];p;p=Next[p])
#define Forpiter(x) for(int &p=iter[x];p;p=Next[p])  
#define Lson (o<<1)
#define Rson ((o<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,127,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define INF (2139062143)
#define F (100000007)
#define pb push_back
#define mp make_pair 
#pragma comment(linker, "/STACK:102400000,102400000")
#define MAXN (200000+10)
typedef long long ll;
ll mul(ll a,ll b){return (a*b)%F;}
ll add(ll a,ll b){return (a+b)%F;}
ll sub(ll a,ll b){return (a-b+llabs(a-b)/F*F+F)%F;}
void upd(ll &a,ll b){a=(a%F+b%F)%F;}

int n,m;
int edge[MAXN],pre[MAXN],Next[MAXN],siz=1;
void addedge(int u,int v)
{
	edge[++siz]=v;
	Next[siz]=pre[u];
	pre[u]=siz;
}
void addedge2(int u,int v){addedge(u,v); addedge(v,u); }

int son[MAXN],top[MAXN];
int fa[MAXN],dep[MAXN],sz[MAXN];
void dfs(int x){
	Forp(x) {
		int v=edge[p];
		if (v==fa[x]) continue;
		fa[v]=x;
		dep[v]=dep[x]+1; 
		dfs(v); 
		sz[x]+=sz[v];
		if (sz[son[x]]<sz[v]) son[x]=v;
	}	
	++sz[x];
} 
int q[MAXN];
void init() {
	MEM(son) MEM(top) MEM(fa) MEM(dep) MEM(sz)
	dfs(1);
	
	int head=1,tail=1;q[1]=1;
	while(head<=tail)
	{
		int now=q[head++];
		if (!top[now]) {
			for(int x=now;x;x=son[x]) top[x]=now;
		} 
		Forp(now) 
		{
			int v=edge[p];
			if (fa[now]!=v) q[++tail]=v; 
		}
	}	
}
int lca(int u,int v) {
	while(1) {
		if (top[u]==top[v]) return dep[u]<dep[v]?u:v;
		if (dep[top[u]]>dep[top[v]]) u=fa[top[u]];
		else v=fa[top[v]];
	}	
} 

bool vis[MAXN];
void color(int x)
{
	vis[x]=1;
	Forp(x) {
		int v=edge[p];
		if (v==fa[x]||vis[v]) continue;
		color(v);	
	}
}
bool check(int u,int v)
{
	return !(vis[u]||vis[v]); 
}

struct chain{
	int u,v,lc;
	chain(){}
	chain(int _u,int _v,int _lc) {u=_u,v=_v,lc=_lc;}
	friend bool operator<(chain a,chain b){return dep[a.lc]>dep[b.lc];	}	
}e[MAXN];

int main()
{
//	freopen("B.in","r",stdin);
//	freopen(".out","w",stdout);
	
	while(scanf("%d%d",&n,&m)==2) {
		
		MEM(edge) MEM(pre) MEM(Next) siz=1;
		MEM(fa) MEM(vis)
		For(i,n-1) {
			int u,v;
			scanf("%d%d",&u,&v);
			addedge2(u,v);
		}
		init();
		
		For(i,m) {
			int u,v;
			scanf("%d%d",&u,&v);
			e[i]=chain(u,v,lca(u,v));
		}
		sort(e+1,e+1+m);
		
		int ans=0;
		For(i,m) {
			if (check(e[i].u,e[i].v)) {
				color(e[i].lc);
				++ans;
			}
		}
		cout<<ans<<endl;
		
	}
	
	
	return 0;
}