题面
给一颗树,每次选一个点,覆盖该点及其周围与其距离不超过k的节点,求最少需要选多少个点。
题解
贪心,每次取深度低的没取的点的K级祖先
显然这样最优
#include<bits/stdc++.h> using namespace std; #define LL long long inline LL read() { LL f = 1,x = 0; char ch; do { ch = getchar(); if(ch == '-') f=-1; }while(ch<'0'||ch>'9'); do { x = (x<<3) + (x<<1) + ch - '0'; ch = getchar(); }while(ch>='0'&&ch<='9'); return f*x; } const int MAXN = 1e5 + 10; int n,k,t; int fa[MAXN]; int head,tail,q[MAXN]; vector<int>ve[MAXN]; int dp[MAXN]; inline void BFS() { head = 1,tail = 1; q[head]=1; while(head<=tail) { int x = q[head];head++; for(int i=0;i<ve[x].size();i++) { int v = ve[x][i]; if(!fa[v]) { fa[v] = x; q[++tail]=v; } } } } inline void color(int now) { if(!dp[now]) return; for(int i=0;i<ve[now].size();i++) { int v = ve[now][i]; if(dp[v]<dp[now]-1) { dp[v]=dp[now]-1; color(v); } } } int main() { n = read(),k = read(),t = read(); for(int i=1;i<n;i++) { int u = read(),v = read(); ve[u].push_back(v); ve[v].push_back(u); } fa[1]=1; BFS(); int ans = 0; memset(dp,-1,sizeof(dp)); reverse(q+1,q+n+1); for(int i=1;i<=n;i++) { if(dp[q[i]]==-1) { ans++; int now = q[i]; for(int j=1;j<=k;j++) now = fa[now]; dp[now]=k; color(now); } } cout<<ans<<endl; }