제출 #98989

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
98989		Benq	Unique Cities (JOI19_ho_t5)	C++14	4 / 100	2060 ms	53248 KiB

joi2019_ho_t5

#pragma GCC optimize ("O3")
#pragma GCC target ("sse4")

#include <bits/stdc++.h>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/rope>

using namespace std;
using namespace __gnu_pbds;
using namespace __gnu_cxx;
 
typedef long long ll;
typedef long double ld;
typedef complex<ld> cd;

typedef pair<int, int> pi;
typedef pair<ll,ll> pl;
typedef pair<ld,ld> pd;

typedef vector<int> vi;
typedef vector<ld> vd;
typedef vector<ll> vl;
typedef vector<pi> vpi;
typedef vector<pl> vpl;
typedef vector<cd> vcd;

template <class T> using Tree = tree<T, null_type, less<T>, rb_tree_tag,tree_order_statistics_node_update>;

#define FOR(i, a, b) for (int i = (a); i < (b); i++)
#define F0R(i, a) for (int i = 0; i < (a); i++)
#define FORd(i,a,b) for (int i = (b)-1; i >= (a); i--)
#define F0Rd(i,a) for (int i = (a)-1; i >= 0; i--)
#define trav(a, x) for (auto& a : x)

#define mp make_pair
#define pb push_back
#define f first
#define s second
#define lb lower_bound
#define ub upper_bound

#define sz(x) (int)x.size()
#define beg(x) x.begin()
#define en(x) x.end()
#define all(x) beg(x), en(x)
#define resz resize

const int MOD = 1000000007; // 998244353
const ll INF = 1e18;
const int MX = 200001;
const ld PI = 4*atan((ld)1);

template<class T> void ckmin(T &a, T b) { a = min(a, b); }
template<class T> void ckmax(T &a, T b) { a = max(a, b); }

namespace input {
    template<class T> void re(complex<T>& x);
    template<class T1, class T2> void re(pair<T1,T2>& p);
    template<class T> void re(vector<T>& a);
    template<class T, size_t SZ> void re(array<T,SZ>& a);

    template<class T> void re(T& x) { cin >> x; }
    void re(double& x) { string t; re(t); x = stod(t); }
    void re(ld& x) { string t; re(t); x = stold(t); }
    template<class Arg, class... Args> void re(Arg& first, Args&... rest) { 
        re(first); re(rest...); 
    }

    template<class T> void re(complex<T>& x) { T a,b; re(a,b); x = cd(a,b); }
    template<class T1, class T2> void re(pair<T1,T2>& p) { re(p.f,p.s); }
    template<class T> void re(vector<T>& a) { F0R(i,sz(a)) re(a[i]); }
    template<class T, size_t SZ> void re(array<T,SZ>& a) { F0R(i,SZ) re(a[i]); }
}

using namespace input;

namespace output {
    template<class T1, class T2> void pr(const pair<T1,T2>& x);
    template<class T, size_t SZ> void pr(const array<T,SZ>& x);
    template<class T> void pr(const vector<T>& x);
    template<class T> void pr(const set<T>& x);
    template<class T1, class T2> void pr(const map<T1,T2>& x);

    template<class T> void pr(const T& x) { cout << x; }
    template<class Arg, class... Args> void pr(const Arg& first, const Args&... rest) { 
        pr(first); pr(rest...); 
    }

    template<class T1, class T2> void pr(const pair<T1,T2>& x) { 
        pr("{",x.f,", ",x.s,"}"); 
    }
    template<class T> void prContain(const T& x) {
        pr("{");
        bool fst = 1; trav(a,x) pr(!fst?", ":"",a), fst = 0; 
        pr("}");
    }
    template<class T, size_t SZ> void pr(const array<T,SZ>& x) { prContain(x); }
    template<class T> void pr(const vector<T>& x) { prContain(x); }
    template<class T> void pr(const set<T>& x) { prContain(x); }
    template<class T1, class T2> void pr(const map<T1,T2>& x) { prContain(x); }
    
    template<class Arg> void ps(const Arg& first) { pr(first,"\n"); } // print w/ spaces
    template<class Arg, class... Args> void ps(const Arg& first, const Args&... rest) { 
        pr(first," "); ps(rest...); 
    }
}

using namespace output;

namespace io {
    void setIn(string s) { freopen(s.c_str(),"r",stdin); }
    void setOut(string s) { freopen(s.c_str(),"w",stdout); }
    void setIO(string s = "") {
        ios_base::sync_with_stdio(0); cin.tie(0); // fast I/O
        if (sz(s)) { setIn(s+".in"), setOut(s+".out"); } // for USACO
    }
}

using namespace io;

namespace modOp {
    int ad(int a, int b, int mod = MOD) { return (a+b)%mod; }
    int sub(int a, int b, int mod = MOD) { return (a-b+mod)%mod; }
    int mul(int a, int b, int mod = MOD) { return (ll)a*b%mod; }
    
    int AD(int& a, int b, int mod = MOD) { return a = ad(a,b,mod); }
    int SUB(int& a, int b, int mod = MOD) { return a = sub(a,b,mod); }
    int MUL(int& a, int b, int mod = MOD) { return a = mul(a,b,mod); }
    
    int po (int b, int p, int mod = MOD) { return !p?1:mul(po(mul(b,b,mod),p/2,mod),p&1?b:1,mod); }
    int inv (int b, int mod = MOD) { return po(b,mod-2,mod); }
    
    int invGeneral(int a, int b) { // 0 < a < b, gcd(a,b) = 1
        if (a == 0) return b == 1 ? 0 : -1;
        int x = invGeneral(b%a,a); 
        return x == -1 ? -1 : ((1-(ll)b*x)/a+b)%b;
    }
}

using namespace modOp;

template<int SZ> struct TreeDiameter {
    int n, dist[SZ], pre[SZ];
    vi adj[SZ];

    void addEdge(int a, int b) {
        adj[a].pb(b), adj[b].pb(a);
    }

    void dfs(int cur) {
        for (int i: adj[cur]) if (i != pre[cur]) {
            pre[i] = cur;
            dist[i] = dist[cur]+1;
            dfs(i);
        }
    }

    void genDist(int cur) {
        memset(dist,0,sizeof dist);
        pre[cur] = -1;
        dfs(cur);
    }

    vi diameter() {
        vi res;
        genDist(1);
        int bes = 0; FOR(i,1,n+1) if (dist[i] > dist[bes]) bes = i;
        res.pb(bes);
        genDist(bes); FOR(i,1,n+1) if (dist[i] > dist[bes]) bes = i;
        res.pb(bes);
        return res;
    }
};


TreeDiameter<MX> T;

int M,CUR,co[MX],c[MX],len[MX];
pi ans[MX][2];
vpi seq;

void modi(int a, int b) {
    a = c[a];
    if (co[a]) CUR --;
    co[a] += b;
    if (co[a]) CUR ++;
}

void solve(int ind, int cur, int dist, int pre) {
    pi maxSub = {-1,-1};
    trav(a,T.adj[cur]) if (a != pre) ckmax(maxSub,{len[a],a});
    int l = 0; trav(a,T.adj[cur]) if (a != pre && a != maxSub.s) ckmax(l,len[a]+1); 
    
    vpi del;
    while (sz(seq) && seq.back().f >= dist-l) {
        del.pb(seq.back());
        modi(seq.back().s,-1);
        seq.pop_back();
    }
    seq.pb({dist,cur}); modi(seq.back().s,1);
    if (maxSub.s != -1) solve(ind,maxSub.s,dist+1,cur);
    modi(seq.back().s,-1);
    seq.pop_back();
    while (sz(seq) && seq.back().f >= dist-(maxSub.f+1)) {
        del.pb(seq.back());
        modi(seq.back().s,-1);
        seq.pop_back();
    }
    ans[cur][ind] = {dist,CUR};
    seq.pb({dist,cur}); modi(seq.back().s,1);
    trav(a,T.adj[cur]) if (a != pre && a != maxSub.s) solve(ind,a,dist+1,cur);
    modi(seq.back().s,-1); seq.pop_back();
    reverse(all(del)); 
    trav(a,del) {
        seq.pb(a);
        modi(seq.back().s,1);
    }
}

void genLen(int x, int y) {
    len[x] = 0;
    trav(i,T.adj[x]) if (i != y) {
        genLen(i,x);
        ckmax(len[x],len[i]+1);
    }
}

int main() {
    // you should actually read the stuff at the bottom
    setIO(); re(T.n,M);
    F0R(i,T.n-1) {
        int a,b; re(a,b);
        T.addEdge(a,b);
    }
    FOR(i,1,T.n+1) re(c[i]);
    auto a = T.diameter();
    
    //ps("WHAT",a);
    
    genLen(a[0],-1);
    solve(0,a[0],0,-1);
    // ps("HUH",sz(seq));
    genLen(a[1],-1);
    solve(1,a[1],0,-1);
    //FOR(i,1,T.n+1) ps("HUH",ans[i][0],ans[i][1]);
    FOR(i,1,T.n+1) ps(max(ans[i][0],ans[i][1]).s);
    // you should actually read the stuff at the bottom
}

/* stuff you should look for
    * int overflow, array bounds
    * special cases (n=1?), set tle
    * do smth instead of nothing and stay organized
*/

컴파일 시 표준 에러 (stderr) 메시지

joi2019_ho_t5.cpp: In function 'void io::setIn(std::__cxx11::string)':
joi2019_ho_t5.cpp:112:35: warning: ignoring return value of 'FILE* freopen(const char*, const char*, FILE*)', declared with attribute warn_unused_result [-Wunused-result]
     void setIn(string s) { freopen(s.c_str(),"r",stdin); }
                            ~~~~~~~^~~~~~~~~~~~~~~~~~~~~
joi2019_ho_t5.cpp: In function 'void io::setOut(std::__cxx11::string)':
joi2019_ho_t5.cpp:113:36: warning: ignoring return value of 'FILE* freopen(const char*, const char*, FILE*)', declared with attribute warn_unused_result [-Wunused-result]
     void setOut(string s) { freopen(s.c_str(),"w",stdout); }
                             ~~~~~~~^~~~~~~~~~~~~~~~~~~~~~

• Subtask 1: 4 / 4
• Subtask 2: 0 / 32
• Subtask 3: 0 / 32
• Subtask 4: 0 / 32