Awesome Shawarma Gym - 101991A(点分治)

Fouad has a raw awesome shawarma, and he is in a city which is represented as an undirected tree. He heard that there is a magical oven that will cook the Shawarma, to make it so delicious. However, In order to acquire the magical oven, there need to be two conditions that should be satisfied in this city:

  • One extra edge that must be added to join two different nodes in the tree (It is allowed to join two nodes, which were previously connected by a direct edge).
  • The number of bridges after adding the new edge should be between [L,R]
  • inclusively.

Please help Fouad to acquire the magical oven, to cook the awesome shawarma by counting in how many ways he can add an edge that satisfies the conditions above.

大意是:给你一棵树,问有多少种方案在加一条边后使图中桥点数量在·[l,r]范围内。

容易想到,树加一条边后会产生一个环,所有不在环上的边都是桥。满足条件的环的长度是[n-r,n-l],也就是求树上要多少距离[n-r-1,n-l-1]的点对,这就是点分治问题了。

无奈不会点分治,只能去学了一波。

暴力的思路,枚举每个点作为根节点,求出所有子节点到它的距离,计算出小于k的距离,然后容斥去重,但是在一条链的时候会T。点分治就是每次求出树的重心,以重心作为根节点。

#include <bits/stdc++.h>

using namespace std;
#define ll long long
const int N = 1e5+123;
vector<int>son[N];
vector<int>path;
int s,root;
ll ans=0;
int vis[N],sum[N],mson[N];
void getroot(int u,int fa){
    sum[u]=1;mson[u]=0;
    for(int i=0;i<(int)son[u].size();i++){
        int v=son[u][i];
        if(v==fa||vis[v])continue;
        getroot(v,u);
        sum[u] += sum[v];
        mson[u]=max(mson[u],sum[v]);
    }
    mson[u]=max(mson[u],s-sum[u]);
    if(mson[u]<mson[root])root=u;
}
void getdis(int u,int fa,int dis){
    path.push_back(dis);
    for(int i=0;i<(int)son[u].size();i++){
        int v=son[u][i];
        if(v==fa||vis[v])continue;
        getdis(v,u,dis+1);
    }
}
ll cal(int u,int dis,int K){
    ll res=0;
    path.clear();
    getdis(u,0,dis);
    sort(path.begin(),path.end());
    for(int i=0,j=path.size()-1;i<j;i++){
        while(path[i]+path[j]>K&&i<j)j--;
        res+=j-i;
    }
    return res;
}
void divide(int u,int K){
    ans += cal(u,0,K);
    vis[u]=1;
    for(int i=0;i<(int)son[u].size();i++){
        int v=son[u][i];
        if(vis[v])continue;
        ans -= cal(v,1,K);
        mson[0]=s=sum[v];root=0;
        getroot(v,0);divide(root,K);

    }
}
ll solve(int n,int K){
    if(K<=0)return 0;
    ans=0;
    memset(vis,0,sizeof(vis));
    mson[0]=s=n;root=0;
    getroot(1,-1);divide(root,K);
    return ans;
}
int main()
{
    //freopen("awesome.in","r",stdin);
    int T,L,R,n,x,y;
    scanf("%d",&T);
    while(T--){
        scanf("%d %d %d",&n, &L,&R);
        for(int i=1;i<=n;i++)son[i].clear();
        for(int i=0;i<n-1;i++){
            scanf("%d %d",&x,&y);
            son[x].push_back(y);
            son[y].push_back(x);
        }
        printf("%lld\n",solve(n,n-L-1)-solve(n,n-R-2));

    }
    return 0;
}

 

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