Paths on the tree

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)

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≤10⁵). 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)

一棵树，给出m条路径，然后问最多选择多少条路径，使得每两条路径之间没有 不论什么公共点，公共边。

把m条路径求出lca，然后依照lca深度从大到小排序，把路径上经过的点标记出来，贪心就可以。

代码：

#include <stdio.h>
#include <iostream>
#include <algorithm>
#include <sstream>
#include <stdlib.h>
#include <string.h>
#include <limits.h>
#include <string>
#include <time.h>
#include <math.h>
#include <queue>
#include <stack>
#include <set>
#include <map>
using namespace std;
#define INF 0x3f3f3f3f
#define eps 1e-8
#define pi acos(-1.0)
typedef long long ll;
const int maxn=100010;
int head[maxn],tol,fa[maxn][20],dep[maxn];
struct Edge{
    int next,to;
    Edge(int _next=0,int _to=0){
        next=_next;to=_to;
    }
}edge[10*maxn];
void addedge(int u,int v){
    edge[tol]=Edge(head[u],v);
    head[u]=tol++;
}
void bfs(int s){
    queue<int> q;
    dep[s]=0,fa[s][0]=s;
    q.push(s);
    while(!q.empty()){
        int u=q.front();
        q.pop();
        for(int i=1;i<20;i++)fa[u][i]=fa[fa[u][i-1]][i-1];
        for(int i=head[u];i!=-1;i=edge[i].next){
            int v=edge[i].to;
            if(v==fa[u][0])continue;
            fa[v][0]=u;
            dep[v]=dep[u]+1;
            q.push(v);
        }
    }
}
int LCA(int x,int y){
    if(dep[x]<dep[y])swap(x,y);
    for(int i=19;i>=0;i--)if((1<<i)&(dep[x]-dep[y]))x=fa[x][i];
    if(x==y)return x;
    for(int i=19;i>=0;i--)if(fa[x][i]!=fa[y][i])x=fa[x][i],y=fa[y][i];
    return fa[x][0];
}
struct QQ{
    int u,v,lca;
}pp[100100];
bool vis[100100];
int n,m;
bool cmp(QQ a,QQ b){
    return dep[a.lca]>dep[b.lca];
}
int main()
{
    while(~scanf("%d%d",&n,&m)){
        memset(head,-1,sizeof(head));
        tol=0;
        for(int i=1;i<n;i++){
            int u,v;
            scanf("%d%d",&u,&v);
            addedge(u,v);
            addedge(v,u);
        }
        bfs(1);
        for(int i=0;i<m;i++){
            scanf("%d%d",&pp[i].u,&pp[i].v);
            pp[i].lca=LCA(pp[i].u,pp[i].v);
        }
        sort(pp,pp+m,cmp);
        memset(vis,0,sizeof(vis));
        int ans=0;
        for(int i=0;i<m;i++){
            int lca=pp[i].lca,flag=1,u=pp[i].u,v=pp[i].v;
            if(vis[u]||vis[v]||vis[lca])continue;
            for(int j=u;j!=lca;j=fa[j][0])
                if(vis[j]){
                    flag=0;break;
                }
            for(int j=v;j!=lca;j=fa[j][0])
                if(vis[j]){
                    flag=0;break;
                }
            if(!flag)continue;
            ans++;
            vis[lca]=1;
            for(int j=u;j!=lca;j=fa[j][0])vis[j]=1;
            for(int j=v;j!=lca;j=fa[j][0])vis[j]=1;
        }
        printf("%d\n",ans);
    }
    return 0;
}