2018-2019 ICPC Northwestern European Regional Programming Contest (NWERC 2018)

目录

• B. Brexit Negotiations
• C. Circuit Board Design

B. Brexit Negotiations

• 得反向建图，正向建图，使用优先队列，在权值相同的情况下，不同的选择会对后续的结果产生影响，而反向建图，从结果出发则可以避免这个影响。
• 以后得习惯逆向思维，反向建图会解决很多问题

#include<bits/stdc++.h>
using namespace std;
 
int n,tot;
const int N=4e5+10;
int head[N],e[N],rd[N];
queue<int>rec;
 
struct Edge{
    int u,v,next;
}E[N];
 
struct node{
    int w,id;
    bool operator < (const node& a) const {
        return w>a.w;
    }
};
priority_queue<node>q;
 
void add(int u,int v)
{
    E[++tot].v=v;
    E[tot].next=head[u];
    head[u]=tot;
}
 
void solve()
{
    for(int i=1;i<=n;++i){
        if(!rd[i]){
            q.push(node{e[i],i});
        }
    }
    //int time=1;
    node now;
    while(!q.empty()){
        now=q.top();
        q.pop();
        rec.push(now.id);
        int u=now.id;
        for(int i=head[u];i;i=E[i].next){
            int v=E[i].v;
            rd[v]--;
            if(!rd[v]) q.push(node{e[v],v});
        }
    }
}
 
 
 
int main()
{
    scanf("%d",&n);
    for(int i=1;i<=n;++i){
        int d;
        scanf("%d%d",&e[i],&d);
        for(int j=0;j<d;++j){
            int u;
            scanf("%d",&u);
            add(i,u);
            rd[u]++;
        }
    }
    solve();
    int ans=0;
    int time=n-1;
    while(!rec.empty()){
        ans=max(e[rec.front()]+time,ans);
        //printf("%d\n",rec.front());
        rec.pop();
        time--;
    }
    printf("%d\n",ans);
    return 0;
}
 
/*
6
2 2 4 3
4 1 5
1 2 2 4
4 1 5
2 0
4 1 3
 
*/

C. Circuit Board Design

• 把圆分成1000份，因为最多只有1000个点，每个点后面有几个子孙，这个点就占圆的千分之几，dfs一遍就行了，队友最开始想的是把同一个圆上的点均分，太复杂了；

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

const double pi = acos(-1);
const int maxn = 1001;
pair<double,double>ans[maxn];
vector<int>g[maxn];
int n, siz[maxn];

void dfs (int u, int fa, double ang)
{
	siz[u]++;
	for (auto v : g[u]) {
		if(v != fa) {
			ans[v].first = ans[u].first + cos(ang);
			ans[v].second = ans[u].second + sin(ang);
			dfs(v,u,ang);
			siz[u] += siz[v];
			ang += pi / (double)(maxn) * (double)(siz[v]);
		}
	}
}



int main()
{
	ios::sync_with_stdio(false);
	cin.tie(0);
	//cin >> n;
	scanf("%d",&n);
	for ( int i = 0; i < n - 1 ; ++i ) {
		int u, v;
		//cin >> u >> v;
		scanf("%d%d",&u,&v);
		g[u].push_back(v);
		g[v].push_back(u);
	}
	dfs(1,-1,0.0);
	for ( int i = 1; i <= n; ++i ) printf("%.10f %.10f\n",ans[i].first,ans[i].second);
	return 0;
}