Police Stations CodeForces - 796D (bfs)
作者:互联网
大意: 给定树, 有k个黑点, 初始满足条件:所有点到最近黑点距离不超过d, 求最多删除多少条边后, 使得原图仍满足条件.
所有黑点开始bfs, 贪心删边.
#include <iostream> #include <algorithm> #include <math.h> #include <cstdio> #include <set> #include <map> #include <string> #include <vector> #include <string.h> #include <queue> #define PER(i,a,n) for(int i=n;i>=a;--i) #define REP(i,a,n) for(int i=a;i<=n;++i) #define hr cout<<'\n' #define pb push_back #define mid ((l+r)>>1) #define lc (o<<1) #define rc (lc|1) #define ls lc,l,mid #define rs rc,mid+1,r #define x first #define y second #define io std::ios::sync_with_stdio(false); #define endl '\n' using namespace std; typedef long long ll; typedef pair<int,int> pii; const int P = 1e9+7; ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;} ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;} void exgcd(ll a,ll b,ll &d,ll &x,ll &y){b?exgcd(b,a%b,d,y,x),y-=a/b*x:x=1,y=0,d=a;} ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;} //head const int N = 4e5+10, INF = 0x3f3f3f3f; int n, k, d, t; int vis[N], vis2[N]; vector<pii> g[N]; int main() { scanf("%d%d%d", &n, &k, &d); REP(i,1,k) scanf("%d", &t), vis[t]=1; REP(i,1,n-1) { int u, v; scanf("%d%d", &u, &v); g[u].pb({v,i}),g[v].pb({u,i}); } queue<int> q; vector<int> ans; REP(i,1,n) if (vis[i]) q.push(i); while (!q.empty()) { int u = q.front();q.pop(); for (auto e:g[u]) { if (vis2[e.y]) continue; if (vis[e.x]) ans.pb(e.y); else q.push(e.x); vis[e.x] = vis2[e.y] = 1; } } printf("%d\n",int(ans.size())); for (int t:ans) printf("%d ",t);hr; }
标签:Police,int,ll,CodeForces,bfs,vis,ans,a%,include 来源: https://www.cnblogs.com/uid001/p/10754330.html