結果
問題 | No.2319 Friends+ |
ユーザー |
|
提出日時 | 2024-11-10 22:41:33 |
言語 | C++23 (gcc 13.3.0 + boost 1.87.0) |
結果 |
AC
|
実行時間 | 566 ms / 3,000 ms |
コード長 | 6,114 bytes |
コンパイル時間 | 5,866 ms |
コンパイル使用メモリ | 320,920 KB |
実行使用メモリ | 101,356 KB |
最終ジャッジ日時 | 2024-11-10 22:42:09 |
合計ジャッジ時間 | 26,410 ms |
ジャッジサーバーID (参考情報) |
judge3 / judge1 |
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ファイルパターン | 結果 |
---|---|
sample | AC * 2 |
other | AC * 45 |
ソースコード
#include <iostream>#include <string>#include <stdio.h>#include <algorithm>#include <vector>#include <tuple>#include <map>#include<set>#include<queue>#include<stack>#include <unordered_set>#include<thread>#include<bits/stdc++.h>#include <numbers>#include <atcoder/all>#include <cstdio>#pragma GCC target("avx")#pragma GCC optimize("O3")#pragma GCC optimize("unroll-loops")//if(a < 0 || h <= a || b < 0 || w <= b)return;using namespace std;using namespace atcoder;using ll = long long;using ld = long double;using ull = unsigned long long;using mint = modint998244353;using mint1 = modint1000000007;//using VL = vector<ll>;template<typename T> using pq = priority_queue<T>;//降順?(最大取り出し)template<typename T> using pqg = priority_queue<T, vector<T>, greater<T>>;//昇順?(最小取り出し)template<typename T> using vector2 = vector<vector<T>>;template<typename T> using vector3 = vector<vector<vector<T>>>;template<typename T> using vector4 = vector<vector<vector<vector<T>>>>;template<typename T> using vector5 = vector<vector<vector<vector<vector<T>>>>>;template<typename T> using vector6 = vector<vector<vector<vector<vector<vector<T>>>>>>;template<typename T> using pairs = pair<T,T>;#define rep(i, n) for (ll i = 0; i < ll(n); i++)#define rep1(i,n) for(int i = 1;i <= int(n);i++)#define repm(i, m, n) for (int i = (m); (i) < int(n);(i)++)#define repmr(i, m, n) for (int i = (m) - 1; (i) >= int(n);(i)--)#define rep0(i,n) for(int i = n - 1;i >= 0;i--)#define rep01(i,n) for(int i = n;i >= 1;i--)// NのK乗根N < 2^64, K <= 64uint64_t kth_root(uint64_t N, uint64_t K) {assert(K >= 1);if (N <= 1 || K == 1) return N;if (K >= 64) return 1;if (N == uint64_t(-1)) --N;auto mul = [&](uint64_t x, uint64_t y) -> uint64_t {if (x < UINT_MAX && y < UINT_MAX) return x * y;if (x == uint64_t(-1) || y == uint64_t(-1)) return uint64_t(-1);return (x <= uint64_t(-1) / y ? x * y : uint64_t(-1));};auto power = [&](uint64_t x, uint64_t k) -> uint64_t {if (k == 0) return 1ULL;uint64_t res = 1ULL;while (k) {if (k & 1) res = mul(res, x);x = mul(x, x);k >>= 1;}return res;};uint64_t res;if (K == 2) res = sqrtl(N) - 1;else if (K == 3) res = cbrt(N) - 1;else res = pow(N, nextafter(1 / double(K), 0));while (power(res + 1, K) <= N) ++res;return res;}// ユークリッドの互除法による最大公約数算出ll GCD(ll a,ll b){if(b == 0)return a;return GCD(b, a % b);}//拡張ユークリッドの互除法による(ax + by = GCD(a,b))を満たすx,yの算出pair<long long, long long> extgcd(long long a, long long b) {if (b == 0) return make_pair(1, 0);long long x, y;tie(y, x) = extgcd(b, a % b);y -= a / b * x;return make_pair(x, y);}struct UnionFind {vector<int> par; // par[i]:iの親の番号 (例) par[3] = 2 : 3の親が2UnionFind(int N) : par(N) { //最初は全てが根であるとして初期化for(int i = 0; i < N; i++) par[i] = i;}int root(int x) { // データxが属する木の根を再帰で得る:root(x) = {xの木の根}if (par[x] == x) return x;return par[x] = root(par[x]);}void unite(int x, int y) { // xとyの木を併合int rx = root(x); //xの根をrxint ry = root(y); //yの根をryif (rx == ry) return; //xとyの根が同じ(=同じ木にある)時はそのままpar[rx] = ry; //xとyの根が同じでない(=同じ木にない)時:xの根rxをyの根ryにつける}bool same(int x, int y) { // 2つのデータx, yが属する木が同じならtrueを返すint rx = root(x);int ry = root(y);return rx == ry;}};ll n;//座標圧縮vector<ll> Ccomp(vector<ll> a){vector<ll> b = a;sort(b.begin(),b.end());b.erase(unique(b.begin(),b.end()),b.end());//ダブり消去vector<ll> rtn;rep(j,a.size()){ll pb = lower_bound(b.begin(),b.end(),a[j]) - b.begin();rtn.push_back(pb);}return rtn;}/// ここから////////////////////////////////////////////using F = ll;using S = ll;string s;ll modPow(ll a, ll n, ll mod) { if(mod==1) return 0;ll ret = 1; ll p = a % mod; while (n) { if (n & 1) ret = ret * p % mod; p = p * p % mod; n >>= 1;} return ret; }void cincout(){ios::sync_with_stdio(false);std::cin.tie(nullptr);cout<< fixed << setprecision(15);}//seg,遅延segの設定-----ここからS op(S a,S b){return a + b;}//何を求めるか(最大値とか)S e(){return 0;}//めんどいやつS mapping (F a,S b){if(a == 0)return b;else return 0;}//遅延処理F composition (F a,F b){return max(a,b);}//遅延中の枝にさらに処理F id(){return 0;}//遅延のモノイドS Op(S a,S b){return max(a,b);}S E(){return 0;}//segここまでstring abc = "abcdefghijklmnopqrstuvwxyz";string Labc = "ABCDEFGHIJKLMNOPQRSTUVWXYZ";ll mod = 998244353;ll mod1 = ll(1e9) + 7;ll INF =ll(1e18);vector<ll> movey = {-1,0,1,0},movex = {0,1,0,-1};bool outc(ll nowy,ll nowx,ll h,ll w){if(nowy < 0 || nowy >= h || nowx < 0 || nowx >= w){return 0;}return 1;}int main() {cincout();ll m;cin >> n >> m;vector<bitset<20000>> here(n);vector<int> place(n);rep(j,n){ll p;cin >> p;p--;here[p][j] = 1;place[j] = p;}vector<bitset<20000>> fri(n);rep(j,m){int a,b;cin >> a >> b;a--,b--;fri[a][b] = 1;fri[b][a] = 1;}ll q;cin >> q;rep(j,q){ll x,y;cin >> x >> y;x--,y--;if(place[x] != place[y] && (here[place[y]] & fri[x]).any()){cout << "Yes" << endl;here[place[x]][x] = 0;place[x] = place[y];here[place[x]][x] = 1;}else cout << "No" << endl;}return 0;}