poj1741 Tree 点分治

Posted 2020-06-23 __560

入门题，算是对树分治有了初步的理解吧。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<vector>
#define REP(i,a,b) for(int i=a;i<=b;i++)
#define MS0(a) memset(a,0,sizeof(a))

using namespace std;

typedef long long ll;
const int maxn=20100;
const int INF=1e9+10;

int n,k;
int u,v,w;
struct Edge
{
    int v,w;
};
vector<Edge> G[maxn];
vector<int> d;
bool vis[maxn];

void dfs_d(int u,int f,int dep)
{
    for(int i=0;i<G[u].size();i++){
        int v=G[u][i].v;
        if(v==f||vis[v]) continue;
        dfs_d(v,u,dep+G[u][i].w);
        d.push_back(dep+G[u][i].w);
    }
}

void dfs_ds(int u,int f,int dep)
{
    d.push_back(dep);
    for(int i=0;i<G[u].size();i++){
        int v=G[u][i].v;
        if(v==f||vis[v]) continue;
        dfs_ds(v,u,dep+G[u][i].w);
    }
}

///---get_root
int rt,balance;
int get_root(int u,int f,int sz)
{
    int cnt=1,balance1=0;
    for(int i=0;i<G[u].size();i++){
        int v=G[u][i].v;
        if(v==f||vis[v]) continue;
        int tmp=get_root(v,u,sz);
        cnt+=tmp;
        balance1=max(balance1,tmp);
    }
    balance1=max(balance1,sz-cnt);
    if(balance1<balance){
        rt=u;
        balance=balance1;
    }
    return cnt;
}

int solve(int u)
{
    rt=u,balance=INF;
    int sz=get_root(u,0,n);
    rt=u,balance=INF;
    get_root(u,0,sz);
    u=rt;
    vis[u]=1;
    int res=0;
    d.clear();
    dfs_d(u,0,0);
    sort(d.begin(),d.end());
    for(int i=0;i<d.size();i++){
        if(d[i]<=k) res++;
    }
    for(int l=0,r=(int)d.size()-1;l<r;l++){
        while(r>l&&d[r]+d[l]>k) r--;
        res+=r-l;
    }
    for(int i=0;i<G[u].size();i++){
        int v=G[u][i].v;
        if(vis[v]) continue;
        d.clear();
        dfs_ds(v,u,G[u][i].w);
        sort(d.begin(),d.end());
        for(int l=0,r=(int)d.size()-1;l<r;l++){
            while(r>l&&d[r]+d[l]>k) r--;
            res-=r-l;
        }
    }
    for(int i=0;i<G[u].size();i++){
        int v=G[u][i].v;
        if(vis[v]) continue;
        res+=solve(v);
    }
    return res;
}

void play()
{
    freopen("in.txt","w",stdout);
    n=10000;k=1000000;
    cout<<n<<" "<<k<<endl;
    for(int i=1;i<n;i++){
        cout<<i<<" "<<i+1<<" "<<1<<endl;
    }
    cout<<0<<" "<<0<<endl;
}

int main()
{
    //play();return 0;
    //freopen("in.txt","r",stdin);
    while(~scanf("%d%d",&n,&k)){
        if(n==0&&k==0) break;
        REP(i,1,n) G[i].clear();
        REP(i,1,n-1){
            scanf("%d%d%d",&u,&v,&w);
            G[u].push_back({v,w});
            G[v].push_back({u,w});
        }
        MS0(vis);
        printf("%d\n",solve(1));
    }
    return 0;
}

View Code

 对于点分治，我的理解就是进行logn次暴力，每次暴力的复杂度为n，总复杂度为n*logn。