Codeforces 337E Book of Evil (DFS)

Topic Link Http://codeforces.com/contest/337/problem/D

Test instructions: to a tree of n nodes, the right side is 1 . . There is a "demon book" on a node in the tree , which

Book let the nodes within the distance of D be affected. m nodes are now known to have received an impact, asking the most

There are a few more nodes available. can you keep the book of demons?

Idea: First we need to find out the affected points, 22 distance from the furthest point, after that, only need

Calculation has The distance between the points and the two points is less than D. So you can find the answer.

find out how to do the farthest point pair: we first take the 1 node as the root node, calculate the depth of each point at this time, and then

after finding the subject The impact of the point, the depth of the largest one node, and then re-calculate the root node with this node

the depth of each point, with To find out at this time The point at which the depth of the impact is the largest of the point v. So this time u,v this

Two points must be the farthest point pair.

#include <iostream> #include <cstdio> #include <cstring> #include <vector>using namespace std; const int Maxn=100010;vector <int> g[maxn];int n,m,d,a[maxn],dp[maxn],b[maxn],c[maxn];void dfs (int u,int fa) {for    (int i=0;i<g[u].size (); i++)        {int v=g[u][i];        if (V==FA) continue;        dp[v]=dp[u]+1;    DFS (V,U);    }}int Main () {int u,v;    scanf ("%d%d%d", &n,&m,&d);    for (int i=0;i<m;i++) scanf ("%d", &a[i]);        for (int i=1;i<n;i++) {scanf ("%d%d", &u,&v);        G[u].push_back (v);    G[v].push_back (U);    } dp[1]=0;    DFS (1,-1);    int s=a[0],t=a[0];    for (int i=1;i<m;i++) if (Dp[s]<dp[a[i]]) s=a[i];    dp[s]=0;    DFS (S,-1);    memcpy (B,dp,sizeof (DP));    for (int i=1;i<m;i++) if (Dp[t]<dp[a[i]]) t=a[i];    dp[t]=0;    DFS (T,-1);    memcpy (C,dp,sizeof (DP));    int ans=0;    for (int i=1;i<=n;i++) if (B[i]<=d && c[i]<=d) ans++; cout<<ans<<enDL;} 

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Codeforces 337E Book of Evil (DFS)