제출 #244207

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
244207		rqi	동기화 (JOI13_synchronization)	C++14	100 / 100	204 ms	9592 KiB

synchronization

#include <bits/stdc++.h>
using namespace std;
 
typedef long long ll;
typedef long double ld;
typedef double db; 
typedef string str; 

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

typedef vector<int> vi; 
typedef vector<ll> vl; 
typedef vector<db> vd; 
typedef vector<str> vs; 
typedef vector<pi> vpi;
typedef vector<pl> vpl; 
typedef vector<pd> vpd; 

#define mp make_pair
#define f first
#define s second
#define sz(x) (int)x.size()
#define all(x) begin(x), end(x)
#define rall(x) (x).rbegin(), (x).rend() 
#define rsz resize
#define ins insert 
#define ft front() 
#define bk back()
#define pf push_front 
#define pb push_back
#define eb emplace_back 
#define lb lower_bound 
#define ub upper_bound 

#define FOR(i,a,b) for (int i = (a); i < (b); ++i)
#define F0R(i,a) FOR(i,0,a)
#define ROF(i,a,b) for (int i = (b)-1; i >= (a); --i)
#define R0F(i,a) ROF(i,0,a)
#define trav(a,x) for (auto& a: x)

const int MOD = 1e9+7; // 998244353;
const int MX = 2e5+5; 
const ll INF = 1e18; 
const ld PI = acos((ld)-1);
const int xd[4] = {1,0,-1,0}, yd[4] = {0,1,0,-1}; 
mt19937 rng((uint32_t)chrono::steady_clock::now().time_since_epoch().count()); 

template<class T> bool ckmin(T& a, const T& b) { 
    return b < a ? a = b, 1 : 0; }
template<class T> bool ckmax(T& a, const T& b) { 
    return a < b ? a = b, 1 : 0; } 
int pct(int x) { return __builtin_popcount(x); } 
int bits(int x) { return 31-__builtin_clz(x); } // floor(log2(x)) 
int cdiv(int a, int b) { return a/b+!(a<0||a%b == 0); } // division of a by b rounded up, assumes b > 0 
int fstTrue(function<bool(int)> f, int lo, int hi) {
    hi ++; assert(lo <= hi); // assuming f is increasing
    while (lo < hi) { // find first index such that f is true 
        int mid = (lo+hi)/2; 
        f(mid) ? hi = mid : lo = mid+1; 
    } 
    return lo;
}

// INPUT
template<class A> void re(complex<A>& c);
template<class A, class B> void re(pair<A,B>& p);
template<class A> void re(vector<A>& v);
template<class A, size_t SZ> void re(array<A,SZ>& a);

template<class T> void re(T& x) { cin >> x; }
void re(db& d) { str t; re(t); d = stod(t); }
void re(ld& d) { str t; re(t); d = stold(t); }
template<class H, class... T> void re(H& h, T&... t) { re(h); re(t...); }

template<class A> void re(complex<A>& c) { A a,b; re(a,b); c = {a,b}; }
template<class A, class B> void re(pair<A,B>& p) { re(p.f,p.s); }
template<class A> void re(vector<A>& x) { trav(a,x) re(a); }
template<class A, size_t SZ> void re(array<A,SZ>& x) { trav(a,x) re(a); }

// TO_STRING
#define ts to_string
str ts(char c) { return str(1,c); }
str ts(bool b) { return b ? "true" : "false"; }
str ts(const char* s) { return (str)s; }
str ts(str s) { return s; }
template<class A> str ts(complex<A> c) { 
    stringstream ss; ss << c; return ss.str(); }
str ts(vector<bool> v) { 
    str res = "{"; F0R(i,sz(v)) res += char('0'+v[i]);
    res += "}"; return res; }
template<size_t SZ> str ts(bitset<SZ> b) {
    str res = ""; F0R(i,SZ) res += char('0'+b[i]);
    return res; }
template<class A, class B> str ts(pair<A,B> p);
template<class T> str ts(T v) { // containers with begin(), end()
    bool fst = 1; str res = "{";
    for (const auto& x: v) {
        if (!fst) res += ", ";
        fst = 0; res += ts(x);
    }
    res += "}"; return res;
}
template<class A, class B> str ts(pair<A,B> p) {
    return "("+ts(p.f)+", "+ts(p.s)+")"; }

// OUTPUT
template<class A> void pr(A x) { cout << ts(x); }
template<class H, class... T> void pr(const H& h, const T&... t) { 
    pr(h); pr(t...); }
void ps() { pr("\n"); } // print w/ spaces
template<class H, class... T> void ps(const H& h, const T&... t) { 
    pr(h); if (sizeof...(t)) pr(" "); ps(t...); }

// DEBUG
void DBG() { cerr << "]" << endl; }
template<class H, class... T> void DBG(H h, T... t) {
    cerr << ts(h); if (sizeof...(t)) cerr << ", ";
    DBG(t...); }
#ifdef LOCAL // compile with -DLOCAL
#define dbg(...) cerr << "LINE(" << __LINE__ << ") -> [" << #__VA_ARGS__ << "]: [", DBG(__VA_ARGS__)
#else
#define dbg(...) 0
#endif

// FILE I/O
void setIn(string s) { freopen(s.c_str(),"r",stdin); }
void setOut(string s) { freopen(s.c_str(),"w",stdout); }
void unsyncIO() { ios_base::sync_with_stdio(0); cin.tie(0); }
void setIO(string s = "") {
    unsyncIO();
    // cin.exceptions(cin.failbit); 
    // throws exception when do smth illegal
    // ex. try to read letter into int
    if (sz(s)) { setIn(s+".in"), setOut(s+".out"); } // for USACO
}

/**
 * Description: Link-Cut Tree. Given a function $f(1\ldots N)\to 1\ldots N,$ 
     * evaluates $f^b(a)$ for any $a,b.$ \texttt{sz} is for path queries; 
     * \texttt{sub}, \texttt{vsub} are for subtree queries. \texttt{x->access()} 
     * brings \texttt{x} to the top and propagates it; its left subtree will be 
     * the path from \texttt{x} to the root and its right subtree will be empty. 
     * Then \texttt{sub} will be the number of nodes in the connected component
     * of \texttt{x} and \texttt{vsub} will be the number of nodes under \texttt{x}.
     * Use \texttt{makeRoot} for arbitrary path queries.
 * Time: O(\log N)
 * Usage: FOR(i,1,N+1)LCT[i]=new snode(i); link(LCT[1],LCT[2],1);
 * Source: Dhruv Rohatgi, Eric Zhang
    * https://sites.google.com/site/kc97ble/container/splay-tree/splaytree-cpp-3
    * https://codeforces.com/blog/entry/67637
 * Verification: (see README for links)
    * ekzhang Balanced Tokens
    * Dynamic Tree Test (Easy)
    * https://probgate.org/viewproblem.php?pid=578 (The Applicant)
 */

typedef struct snode* sn;
struct snode { //////// VARIABLES
    sn p, c[2]; // parent, children
    sn extra; // extra cycle node for "The Applicant"
    bool flip = 0; // subtree flipped or not
    int val, sz; // value in node, # nodes in current splay tree
    int sub, vsub = 0; // vsub stores sum of virtual children
    snode(int _val) : val(_val) {
        p = c[0] = c[1] = extra = NULL; calc(); }
    friend int getSz(sn x) { return x?x->sz:0; }
    friend int getSub(sn x) { return x?x->sub:0; }
    void prop() { // lazy prop
        if (!flip) return;
        swap(c[0],c[1]); flip = 0;
        F0R(i,2) if (c[i]) c[i]->flip ^= 1;
    }
    void calc() { // recalc vals
        F0R(i,2) if (c[i]) c[i]->prop();
        sz = 1+getSz(c[0])+getSz(c[1]);
        sub = 1+getSub(c[0])+getSub(c[1])+vsub;
    }
    //////// SPLAY TREE OPERATIONS
    int dir() {
        if (!p) return -2;
        F0R(i,2) if (p->c[i] == this) return i;
        return -1; // p is path-parent pointer
    } // -> not in current splay tree
    // test if root of current splay tree
    bool isRoot() { return dir() < 0; } 
    friend void setLink(sn x, sn y, int d) {
        if (y) y->p = x;
        if (d >= 0) x->c[d] = y; }
    void rot() { // assume p and p->p propagated
        assert(!isRoot()); int x = dir(); sn pa = p;
        setLink(pa->p, this, pa->dir());
        setLink(pa, c[x^1], x); setLink(this, pa, x^1);
        pa->calc(); calc();
    }
    void splay() {
        while (!isRoot() && !p->isRoot()) {
            p->p->prop(), p->prop(), prop();
            dir() == p->dir() ? p->rot() : rot();
            rot();
        }
        if (!isRoot()) p->prop(), prop(), rot();
        prop();
    }
    sn fbo(int b) { // find by order
        prop(); int z = getSz(c[0]); // of splay tree
        if (b == z) { splay(); return this; }
        return b < z ? c[0]->fbo(b) : c[1] -> fbo(b-z-1);
    }
    //////// BASE OPERATIONS
    void access() { // bring this to top of tree, propagate
        for (sn v = this, pre = NULL; v; v = v->p) {
            v->splay(); // now switch virtual children
            if (pre) v->vsub -= pre->sub;
            if (v->c[1]) v->vsub += v->c[1]->sub;
            v->c[1] = pre; v->calc(); pre = v;
        }
        splay(); assert(!c[1]); // right subtree is empty
    }
    void makeRoot() { 
        access(); flip ^= 1; access(); assert(!c[0] && !c[1]); }
    //////// QUERIES
    friend sn lca(sn x, sn y) {
        if (x == y) return x;
        x->access(), y->access(); if (!x->p) return NULL;
        x->splay(); return x->p?:x; // y was below x in latter case
    } // access at y did not affect x -> not connected
    friend bool connected(sn x, sn y) { return lca(x,y); } 
    // # nodes above
    int distRoot() { access(); return getSz(c[0]); } 
    sn getRoot() { // get root of LCT component
        access(); sn a = this; 
        while (a->c[0]) a = a->c[0], a->prop();
        a->access(); return a;
    }
    sn getPar(int b) { // get b-th parent on path to root
        access(); b = getSz(c[0])-b; assert(b >= 0);
        return fbo(b);
    } // can also get min, max on path to root, etc
    //////// MODIFICATIONS
    void set(int v) { access(); val = v; calc(); } 
    friend void link(sn x, sn y, bool force = 0) { 
        assert(!connected(x,y)); 
        if (force) y->makeRoot(); // make x par of y
        else { y->access(); assert(!y->c[0]); }
        x->access(); setLink(y,x,0); y->calc();
    }
    friend void cut(sn y) { // cut y from its parent
        y->access(); assert(y->c[0]);
        y->c[0]->p = NULL; y->c[0] = NULL; y->calc(); }
    friend void cut(sn x, sn y) { // if x, y adj in tree
        y->access();
        if(y->c[0] != x){
            cut(y, x);
            return;
        }
        cut(y); }
};

int N, M, Q;
sn LCT[MX];
bool on[MX];
int sum[MX];
int X[MX];
int Y[MX];
int ans[MX];

int main() {
    setIO();
    
    cin >> N >> M >> Q;
    for(int i = 1; i <= N-1; i++){
        cin >> X[i] >> Y[i];
    }

    for(int i = 1; i <= N; i++){
        LCT[i] = new snode(1);
    }

    for(int i = 1; i <= M; i++){
        int D;
        cin >> D;
        on[D]^=1;
        if(on[D]){ //turn on
            sn a = LCT[X[D]];
            sn b = LCT[Y[D]];
            a = a->getRoot();
            b = b->getRoot();
            int num1 = a->val;
            int num2 = b->val;
            int num3 = sum[D];
            int num = num1+num2-num3;
            link(LCT[X[D]], LCT[Y[D]]);
            (a->getRoot())->val = num;
        }   
        else{ //turn off
            sn a = LCT[X[D]];
            sn b = LCT[Y[D]];
            int num = (a->getRoot())->val;
            cut(a, b);
            sum[D] = num;
            (a->getRoot())->val = num;
            (b->getRoot())->val = num;
        }
    }

    for(int i = 1; i <= N; i++){
        ans[i] = (LCT[i]->getRoot())->val;
    }

    for(int i = 1; i <= Q; i++){
        int C;
        cin >> C;
        ps(ans[C]);
    }

    // you should actually read the stuff at the bottom
}

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

컴파일 시 표준 에러 (stderr) 메시지

synchronization.cpp: In function 'void setIn(std::__cxx11::string)':
synchronization.cpp:128:31: 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); }
                        ~~~~~~~^~~~~~~~~~~~~~~~~~~~~
synchronization.cpp: In function 'void setOut(std::__cxx11::string)':
synchronization.cpp:129:32: 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: 10 / 10
• Subtask 2: 20 / 20
• Subtask 3: 10 / 10
• Subtask 4: 20 / 20
• Subtask 5: 40 / 40