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  • Police Stations CodeForces

    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<<'
    '
    #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 '
    '
    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
    ",int(ans.size()));
    	for (int t:ans) printf("%d ",t);hr;
    }
    
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  • 原文地址:https://www.cnblogs.com/uid001/p/10754330.html
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