#1175384 (C++23) No.3617 Swap

• 問題: No.3617 Swap
• ユーザー: yuusaan
• 提出日時: 2026-08-06 15:44:19
• 言語: C++23 (gcc 15.2.0 + boost 1.90.0)
• 結果: WA
• 実行時間: -
• コード長: 7,129 bytes
• コンパイル時間: 4,971 ms
• コンパイル使用メモリ: 395,212 KB
• 実行使用メモリ: 112,964 KB
• 最終ジャッジ日時: 2026-08-06 15:44:39
• 合計ジャッジ時間: 18,009 ms
• ジャッジサーバーID （参考情報）: judge1_1 / judge3_0

サブタスク	配点	結果
サンプル	0 %	AC * 3 WA * 1
小課題1	5 %	AC * 3
小課題2	5 %	AC * 5
小課題3	15 %	AC * 14
小課題4	20 %	AC * 21
小課題5	25 %	WA * 8
小課題6	10 %	WA * 8
小課題7	20 %	AC * 29 WA * 27
合計	3 * 45% = 135 点

ソースコード

//眠いので続きはあしたやります

#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<chrono>
#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 mint = modint1000000007;
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(ll i = 1;i <= ll(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 <= 64
uint64_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 {
  int n;
  vector<int> parent;
  vector<int> size;

  UnionFind(int n) : n(n), parent(n), size(n, 1) {
    for (int i = 0; i < n; i++) {
      parent[i] = i;
    }
  }

  int root(int x) {
    if (parent[x] == x) return x;
    return parent[x] = root(parent[x]);
  }

  bool unite(int x, int y) {
    x = root(x);
    y = root(y);
    if (x == y) return false;
    if (size[x] < size[y]) swap(x, y);
    parent[y] = x;
    size[x] += size[y];
    return true;
  }

  bool same(int x, int y) { return root(x) == root(y); }
};

//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;
}

string abc = "abcdefghijklmnopqrstuvwxyz";
string Labc = "ABCDEFGHIJKLMNOPQRSTUVWXYZ";

ll INF = ll((1LL<<61)+1);
ll mod = 998244353;
ll mod1 = ll(1e9) + 7;

int inf = int(1e9+10);
vector<ll> movey = {-1,0,1,0,-1,-1,1,1},movex = {0,1,0,-1,-1,1,-1,1};
bool outc(ll nowy,ll nowx,ll h,ll w){
    if(nowy < 0 || nowy >= h || nowx < 0 || nowx >= w){
        return 0;
    }
    return 1;
}

vector<mint> vecfact(2,1);
mint fact(ll x){
    if(x >= 500 + vecfact.size()){
        fact(x-500);
    }
    if(x >= vecfact.size()){
        mint get = fact(x-1);
        vecfact.push_back(get * x);
    }
    return vecfact[x];
}

ll maxim = 200000;
vector<ll> bitprimes = {999999929,999999937,1000000021,1000000033,1000000087,1000000093,1000000097};
using mint9 = modint;
ll mod9 = bitprimes[rand() % 7];
/// ここから////////////////////////////////////////////




using F = ll;
using S = array<ll,10>;
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){
    array<ll,20> c;
    rep(j,10)c[j] = a[j];
    rep(j,10)c[10+j] = b[j];
    sort(c.begin(),c.end());
    array<ll,10> d;
    rep(j,10)d[j] = c[j];
    return d;
}

S e(){
    array<ll,10> rt;
    rep(j,10)rt[j] = INF;
    return rt;
};

// S opb(S a,S b){
//     return max(a,b);
// }
// S eb(){return -INF;};

// S mapping (F a,S b){
//     return a + b;
// }//遅延処理

// F composition (F a,F b){
//     return a + b;
// }//遅延中の枝にさらに処理

// F id(){return 0;}//遅延のモノイド

// using LST= lazy_segtree<S,op,e,F,mapping,composition,id>;
using ST = segtree<S,op,e>;

//segここまで
mint ncr(ll n,ll r){
    return fact(n)/fact(r)/fact(n-r);
}

void solve() {
    ll n,m,q;
    cin >> n >> m >> q;
    vector<array<ll,2>> lr(m);
    rep(j,m){
        cin >> lr[j][0] >> lr[j][1];
        lr[j][0]--,lr[j][1]--;
    }
    vector<ll> a(n);
    rep(j,n)a[j] = j;
    vector<array<ll,2>> t(q);
    rep(j,q){
        cin >> t[j][0] >> t[j][1];
        t[j][1]--;
    }
    rep(j,m){
        swap(a[lr[j][0]],a[lr[j][1]]);
    }
    vector b(61,vector<ll>(n));
    vector d(61,vector<ll>(n));
    rep(j,n)b[0][j] = a[j];
    rep1(j,60){
        rep(i,n){
            b[j][i] = b[j-1][b[j-1][i]];
            d[j][b[j][i]] = i;
        }
    }
    rep(j,q){
        ll get = t[j][0]/m;
        rep(i,61){
            if(get&(1ll<<i)){
                t[j][1] = b[i][t[j][1]];
            }
        }
    }
    vector<array<ll,3>> tt(q);
    rep(j,q){
        tt[j][0] = t[j][0]%m;
        tt[j][1] = t[j][1];
        tt[j][2] = j;
    }
    sort(tt.begin(),tt.end());
    ll now = 0;
    vector<ll> ans(q);
    rep(j,n)a[j] = j;
    rep(j,q){
        while(now<tt[j][0]){
            swap(a[lr[now][0]],a[lr[now][1]]);
            now++;
        }
        ans[tt[j][2]] = a[tt[j][1]];
    }
    rep(j,q){
        cout << ans[j]+1 << endl;
    }
    return;
}

signed main() {
    cincout();
    ll t;
    t = 1;
    // cin >> t;
    rep(tt,t)solve();
    return 0;
};