csp-s模拟12

又双叒叕垫底啦！！！

rank 22，T1 0，T2 20，T3 0，T4 30。

逆天模拟赛，逆天题面，怕你赛场上不会打暴力真是煞费苦心出了一场暴力专场。

小h的几何

高斯消元求圆心精度被卡炸了，乐了。咋还考向量啊，不学文化课，输了。

九点圆的圆心是外心\(O\)和垂心\(H\)的中点，且\(\overrightarrow{OH}=\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}\)，因为\(O\)为原点，所以\(\Omega(A,B,C)=\frac{A+B+C}{2}\)。

证明的话不好写，机房没有条件，画图太难受的。可以百度一下。

用到向量的相关知识，文化课好题。

upd:发现有大佬写了，挂一下。come from DrRatio。

点此查看代码

#include<bits/stdc++.h>
#include<bits/extc++.h>
// using namespace __gnu_pbds;
// using namespace __gnu_cxx;
using namespace std;
#define rep(i,s,t,p) for(int i = s;i <= t; i += p)
#define drep(i,s,t,p) for(int i = s;i >= t; i -= p)
#define Infile(x) freopen(#x".in","r",stdin)
#define Outfile(x) freopen(#x".out","w",stdout)
#define Ansfile(x) freopen(#x".ans","w",stdout)
#define Errfile(x) freopen(#x".err","w",stderr)
#ifdef LOCAL
    FILE *InFile = Infile(in),*OutFile = Outfile(out);
    FILE *ErrFile = Errfile(err);
#else
    FILE *InFile = Infile(geometry),*OutFile = Outfile(geometry);
#endif
using ll=long long;using ull=unsigned long long;
using db = double;using ldb = long double;
const int N = 5e5 + 10;
const ldb pi = 3.1415926535897932384626;
struct node{ldb x,y;}p[N];
int n,ct[N];
ldb res,ansx,ansy;
inline void solve(){
    cin>>n;
    res = 3.0/n/(n-1)/(n-2);
    rep(i,1,n,1){
        int q;cin>>q;
        ldb ct = 1.0*q/1e9*pi;
        p[i].x = cos(ct),p[i].y = sin(ct);
    }
    ldb ansx = 0,ansy = 0;
    ll tot = 1ll*(n-1)*(n-2)/2;
    rep(i,1,n,1){
        ansx += tot*p[i].x*res;
        ansy += tot*p[i].y*res;
    }
    cout<<fixed<<setprecision(16)<<ansx<<' '<<ansy<<'\n';
}
signed main(){
    cin.tie(nullptr)->sync_with_stdio(false);
    solve();
}

小w的代数

想等学了圆方树以后去打\(O(n^2)\)的做法，仙姑了。

小y的数论

难题。

膜拜神已经要切了。

小j的组合

简单题，没场切挺遗憾的。

考虑求出树的直径以后，答案\(g=n-|树的直径|\)。

证明的话，考虑贪心一下，跑dfs，如果要走哈密顿路的话最后肯定是一条链。

考虑到如果要让新增的节点数最少的话肯定要让dfs回溯所经过边的最少，那么肯定是除了树的直径以外的其余边回溯，直接求树的直径即可。

然后求树的直径，暴力建边，dfs求哈密顿路，注意直径上的点要最后走。

时间复杂度\(O(n+m)\)

点此查看代码

#include<bits/stdc++.h>
#include<bits/extc++.h>
// using namespace __gnu_pbds;
// using namespace __gnu_cxx;
using namespace std;
#define rep(i,s,t,p) for(int i = s;i <= t; i += p)
#define drep(i,s,t,p) for(int i = s;i >= t; i -= p)
#define Infile(x) freopen(#x".in","r",stdin)
#define Outfile(x) freopen(#x".out","w",stdout)
#define Ansfile(x) freopen(#x".ans","w",stdout)
#define Errfile(x) freopen(#x".err","w",stderr)
#ifdef LOCAL
    FILE *InFile = Infile(in),*OutFile = Outfile(out);
    // FILE *ErrFile = Errfile(err);
#else
    FILE *InFile = Infile(combo),*OutFile = Outfile(combo);
#endif
using ll=long long;using ull=unsigned long long;
using db = double;using ldb = long double;
#define eb emplace_back
const int N = 210;
vector<int> e[N];
int n,s,t,dep[N],fa[N],siz[N],son[N],ed[N],bf[N];
bitset<N> vis,flag;
inline int bfs(int s){
    queue<int> q;q.push(s);
    rep(i,1,n,1) dep[i] = bf[i] = 0;
    vis.reset();bf[s] = 0;dep[s] = 1;
    int mxdep = 0,far = s;
    while(q.size()){
        int x = q.front();q.pop();vis[x] = true;
        for(int y:e[x]){
            if(vis[y]) continue;
            dep[y] = dep[x] + 1;
            if(dep[y] > mxdep){mxdep = dep[y],far = y;}
            bf[y] = x;
            q.push(y);
        }
    }
    return far;
}
vector<int> ans;
inline void Bfs(int x){
    queue<int> q;q.push(x);
    while(q.size()){
        int x = q.front();q.pop();vis[x] = true;
        for(int y:e[x]){
            if(flag[y] || vis[y]) continue;
            q.push(y);ans.eb(x);
        }
    }
}
vector<int> out;
void dfs(int x,int f){
    out.eb(x);
    vis[x] = true;
    int to = 0;
    for(int y:e[x]){
        if(vis[y]) continue;
        if(flag[y]){to = y;continue;}
        dfs(y,x);
    }
    if(to && !vis[to]) dfs(to,x);
}
inline void solve(){
    cin>>n;rep(i,1,n-1,1){int u,v;cin>>u>>v;e[u].eb(v),e[v].eb(u);}
    s = bfs(t = bfs(1));
    int x = s;vis.reset();
    do{flag.set(x),x = bf[x];}while(x);
    rep(i,1,n,1) if(flag[i]) Bfs(i);
    int now = n+1;
    for(auto x:ans){
        for(int y:e[x]) e[y].eb(now),e[now].eb(y);
        now++;
    }now--;
    vis.reset();dfs(s,0);
    cout<<ans.size()<<'\n';
    for(auto i:ans) cout<<i<<' ';
    cout<<'\n';
    for(auto i:out) cout<<i<<' ';
}
signed main(){
    cin.tie(nullptr)->sync_with_stdio(false);
    solve();
}

附赠spj

点此查看代码

#include<bits/stdc++.h>
using namespace std;
#define rep(i,s,t,p) for(int i = s;i <= t; i += p)
#define drep(i,s,t,p) for(int i = s;i >= t; i -= p)
using ll=long long;using ull=unsigned long long;
using db = double;using ldb = long double;
const int N = 1e5 + 10;
vector<int> e[N];
#define eb emplace_back
int n,ansg,outg,len;
vector<int> dis;
bitset<N> have;
bool dfs(int now){
    if(now == len-1) return true;
    bool flag = true;
    for(int y:e[dis[now]]){
        if(y == dis[now + 1]){
            return dfs(now + 1);
        }
    }
    return false;
}
signed main(){
    ifstream cin("in.in");ofstream cout("spj.out");
    cin>>n;
    rep(i,1,n-1,1){int u,v;cin>>u>>v;e[u].eb(v);e[v].eb(u);}
    cin.close();cin.open("ans.ans");
    cin>>ansg;cin.close();cin.open("out.out");
    cin>>outg;if(ansg != outg){
        cout<<"What Are You Fucking\n";
        return 0;
    }
    int now = n;
    rep(i,1,outg,1){
        int x;cin>>x;now++;
        for(int y:e[x]) e[y].eb(now),e[now].eb(y);
    }
    rep(i,1,now,1){
        int x;cin>>x;
        if(have[x]) return cout<<"You Are A Joker\n",0;
        have.set(x),dis.eb(x);
    }
    if(have.count() != now){return cout<<"You Are A Joker\n",0;}
    len = dis.size();
    cout<<(dfs(0)?"TMD，终于AC了，你个shaber\n":"You Are A Joker\n");   
}