D. Book of Evil

合集 - 树论(34)

1.P3478 [POI2008] STA-Station2024-03-192.E1. Weights Division (easy version)2024-03-223.P5007 DDOSvoid 的疑惑2024-03-254.P2168 [NOI2015] 荷马史诗2024-03-305.P8306 【模板】字典树2024-03-306.P1481 魔族密码2024-03-307.P3128 [USACO15DEC] Max Flow P2024-04-028.P3258 [JLOI2014] 松鼠的新家2024-04-029.P5536 【XR-3】核心城市2024-04-0210.P5836 [USACO19DEC] Milk Visits S2024-04-0211.P3384 【模板】重链剖分/树链剖分2024-04-0312.P3038 [USACO11DEC] Grass Planting G2024-04-0313.P4551 最长异或路径2024-04-0414.F. Alex's whims2024-04-1215.J. 上学2024-05-1416.Game on Tree2024-05-1417.E. We Need More Bosses2024-05-1518.A. Party2024-05-1519.D. Strong Vertices2024-05-1620.B. Omkar and Heavenly Tree2024-05-1621.B. Mahmoud and Ehab and the bipartiteness2024-05-1622.P8655 [蓝桥杯 2017 国 B] 发现环2024-05-2423.P10298 [CCC 2024 S4] Painting Roads2024-05-2424.F. Gardening Friends2024-06-1425.D. Playoff Tournament2024-06-1926.E. Lomsat gelral2024-07-1227.F. Minimum Maximum Distance2024-07-14

28.D. Book of Evil2024-07-15

29.P2195 HXY造公园2024-07-1530.P3304 [SDOI2013] 直径2024-07-1531.D. Sasha and a Walk in the City2024-07-1932.D. Vitaly and Cycle2024-08-0133.P10838 『FLA - I』庭中有奇树2024-08-0734.F - Perfect Matching on a Tree2024-08-10

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原题链接

题解

题目要求有多少个点，其到标记点的最远距离不超过 $d$

看到这个我们不难想到树的直径：设直径端点 $a,b$，树上任意一点 $c$ 到叶子节点的距离 $\le max(d(c,a),d(c,b))$

所以，我们把标记点看成叶子节点，并找出相距最远的一对标记点 $ab$，如果某点与 $a,b$ 的距离都不超过 $d$ 则与其他标记点的距离也不会超过 $d$

(如果有非标记点，在叶子节点的“外侧”，该性质依然成立)

code

#include<bits/stdc++.h>
#define ll long long
using namespace std;

vector<int> G[100004];
bool mark[100005]={0};

int maxh=0;
int node1;
int node2;

void dfs(int now,int fa,int height)
{

    if(height>maxh&&mark[now])
    {
        maxh=height;
        node1=now;
    }
    for(auto next:G[now])
    {
        if(next==fa) continue;
        dfs(next,now,height+1);
    }
}

int ans[100005]={0};
int n,m,d;

void dfs1(int now,int fa,int height)
{
    if(height<=d) ans[now]++;
    if(height>maxh&&mark[now])
    {
        maxh=height;
        node2=now;
    }
    for(auto next:G[now])
    {
        if(next==fa) continue;
        dfs1(next,now,height+1);
    }
}

void dfs2(int now,int fa,int height)
{
    if(height<=d) ans[now]++;

    for(auto next:G[now])
    {
        if(next==fa) continue;
        dfs2(next,now,height+1);
    }
}


void solve()
{
    cin>>n>>m>>d;

    int start;
    for(int i=1;i<=m;i++)
    {
        int x;
        cin>>x;
        mark[x]=1;
        start=x;
    }

    for(int i=1;i<n;i++)
    {
        int x,y;
        cin>>x>>y;
        G[x].push_back(y);
        G[y].push_back(x);
    }

    node1=start;
    dfs(start,start,0);

    maxh=0;
    node2=start;
    dfs1(node1,node1,0);

    dfs2(node2,node2,0);

    int sum=0;
    for(int i=1;i<=n;i++) sum+=(ans[i]==2);
    cout<<sum<<'\n';
}
int main()
{
    ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
    int t=1;
    //cin>>t;
    while(t--) solve();
    return 0;
}