Paths on the tree

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1097    Accepted Submission(s): 366

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)

Source

2014 Multi-University Training Contest 5

从最深的公共祖先开始选择，选完之后，他的所有子节点都不可能再选，不然就重复了

用倍增思想求解即可

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<cstdlib>
#include<string>
#include<algorithm>
#include<queue>
#include<vector>
#include<stack>
#include<map>
#include<set>
using namespace std;
vector<int> e[100005];
struct node
{
    int x,y,z;
}mp[100005];
int n,m,vis[100005],deep[100005],f[100005][30],mark[100005];
void init()
{
    for(int i=1;i<=n;i++)
    {
        vis[i]=0;
        deep[i]=0;
        mark[i]=0;
        e[i].clear();
    }
    memset(f,0,sizeof(f));
}
void dfs(int u)
{
    vis[u]=1;
    for(int i=0;i<e[u].size();i++)
    {
        int v=e[u][i];
        if(v==f[u][0])
            continue;
        if(!vis[v])
        {
            deep[v]=deep[u]+1;
            f[v][0]=u;
            dfs(v);
        }
    }
}
void bz()
{
    for(int i=1;(1<<i)<=n;i++)
    {
        for(int j=1;j<=n;j++)
        {
            f[j][i]=f[f[j][i-1]][i-1];
        }
    }
}
bool cmp(node a,node b)
{
    return deep[a.z]>deep[b.z];
}
int LCA(int x,int y)
{
    int i,j;
    if(deep[x]<deep[y])
        swap(x,y);
    for(i=0;(1<<i)<=deep[x];i++);
    i--;
    for(j=i;j>=0;j--)
    {
        if(deep[x]-(1<<j)>=deep[y])
        {
            x=f[x][j];
        }
    }
    if(x==y)
        return y;
    for(j=i;j>=0;j--)
    {
        if(f[x][j]!=f[y][j])
        {
            x=f[x][j];
            y=f[y][j];
        }
    }
    return f[x][0];
}
void col(int u)
{
    mark[u]=1;
    for(int i=0;i<e[u].size();i++)
    {
        int v=e[u][i];
        if(deep[v]==deep[u]+1&&!mark[v])
            col(v);
    }
}
int main()
{
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        init();
        for(int i=1;i<n;i++)
        {
            int x,y;
            scanf("%d%d",&x,&y);
            e[x].push_back(y);
            e[y].push_back(x);
        }
        f[1][0]=1;
        deep[1]=1;
        dfs(1);
        bz();
        for(int i=0;i<m;i++)
        {
            int x,y;
            scanf("%d%d",&x,&y);
            mp[i].x=x,mp[i].y=y,mp[i].z=LCA(x,y);
        }
        int ans=0;
        sort(mp,mp+m,cmp);
        for(int i=0;i<m;i++)
        {
            if(!mark[mp[i].x]&&!mark[mp[i].y])
            {
                ans++;
                col(mp[i].z);
            }
        }
        printf("%d\n",ans);
    }
    return 0;
}