結果

問題 No.3072 Speedrun Query
ユーザー The Forsaking
提出日時 2025-03-24 18:16:23
言語 C++17
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 164 ms / 2,500 ms
コード長 3,225 bytes
コンパイル時間 1,223 ms
コンパイル使用メモリ 116,760 KB
実行使用メモリ 16,560 KB
最終ジャッジ日時 2025-03-24 18:16:30
合計ジャッジ時間 6,695 ms
ジャッジサーバーID
（参考情報） 		judge1 / judge3
このコードへのチャレンジ
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ファイルパターン 結果
sample AC * 1
other AC * 21
権限があれば一括ダウンロードができます
コンパイルメッセージ
main.cpp: In function ‘int main()’:
main.cpp:64:14: warning: ignoring return value of ‘int scanf(const char*, ...)’ declared with attribute ‘warn_unused_result’ [-Wunused-result]
   64 |         scanf("%d", &x);
      |         ~~~~~^~~~~~~~~~
main.cpp:68:14: warning: ignoring return value of ‘int scanf(const char*, ...)’ declared with attribute ‘warn_unused_result’ [-Wunused-result]
   68 |         scanf("%d", &x);
      |         ~~~~~^~~~~~~~~~
main.cpp:100:14: warning: ignoring return value of ‘int scanf(const char*, ...)’ declared with attribute ‘warn_unused_result’ [-Wunused-result]
  100 |         scanf("%d%d", &x, &y);
      |         ~~~~~^~~~~~~~~~~~~~~~

ソースコード

diff #
プレゼンテーションモードにする

#include <iostream>
#include <sstream>
#include <cstring>
#include <string>
#include <algorithm>
#include <cmath>
#include <map>
#include <set>
#include <vector>
#include <queue> 
#include <unordered_set>
#include <unordered_map>
#include <bitset>
#include <ctime>
#include <assert.h>
#include <deque>
#include <list>
#include <stack>
#include <numeric>
#include <iomanip>
using namespace std;
 
typedef pair<long long, int> pli;
typedef pair<int, long long> pil;
typedef pair<long long , long long> pll;
typedef pair<int, int> pii;
typedef pair<double, double> pdd;
typedef pair<int, pii> piii;
typedef pair<int, long long > pil;
typedef pair<long long, pii> plii;
typedef pair<double, int> pdi;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<ull, ull> puu;
typedef long double ld;
const int N = 2000086, MOD = 998244353, INF = 0x3f3f3f3f, MID = 333;
const long double EPS = 1e-8;
int dx[4] = {1, 0, -1, 0}, dy[4] = {0, 1, 0, -1};
// int dx[8] = {1, 1, 0, -1, -1, -1, 0, 1}, dy[8] = {0, 1, 1, 1, 0, -1, -1, -1};
// int dx[8] = {2, 1, -1, -2, -2, -1, 1, 2}, dy[8] = {1, 2, 2, 1, -1, -2, -2, -1};
int n, m, cnt;
int w[N];
vector<ll> num;
ll res;
ll lowbit(ll x) { return x & -x; }
ll lcm(ll a, ll b) { return a / __gcd(a, b) * b; }
inline double rand(double l, double r) { return (double)rand() / RAND_MAX * (r - l) + l; }
inline ll qmi(ll a, ll b, ll c) { ll res = 1; while (b) { if (b & 1) res = res * a % c; a = a * a % c; b >>= 1; } return res; }
inline ll qmi(ll a, ll b) { ll res = 1; while (b) { if (b & 1) res *= a; a *= a; b >>= 1; } return res; }
inline double qmi(double a, ll b) { double res = 1; while (b) { if (b & 1) res *= a; a *= a; b >>= 1; } return res; } 
inline ll C(ll a, ll b, int* c) { if (a < b) return 0; ll res = 1; for (ll j = a, i = 1; i < b + 1; i++, j--) res *= j; for (ll j = a, i = 1; i < b + 
    1; i++, j--) res /= i; return res; }
inline int find_(int x) { return lower_bound(num.begin(), num.end(), x) - num.begin(); }
bool st[2][N];
int d[2][N];
int main() {
    cin >> n;
    int ka, kb, x;
    cin >> ka >> kb;
    while (ka--) {
        scanf("%d", &x);
        st[0][x] = 1;
    }
    while (kb--) {
        scanf("%d", &x);
        st[1][x] = 1;
    }
    for (int i = 1; i < n + 1; i++) d[0][i] = d[1][i] = INF;
    int p = -INF;
    for (int i = 1; i < n + 1; i++) {
        if (st[0][i]) p = i;
        d[0][i] = min(d[0][i], i - p);
    }
    p = -INF;
    for (int i = 1; i < n + 1; i++) {
        if (st[1][i]) p = i;
        d[1][i] = min(d[1][i], i - p);
    }
    p = INF;
    for (int i = n; i; i--) {
        if (st[0][i]) p = i;
        d[0][i] = min(d[0][i], p - i);
    }
    p = INF;
    for (int i = n; i; i--) {
        if (st[1][i]) p = i;
        d[1][i] = min(d[1][i], p - i);
    }
    int dab = INF;
    for (int i = 1; i < n + 1; i++)
        if (st[0][i])
            dab = min(dab, d[1][i]);
    cin >> m;
    while (m--) {
        int x, y;
        scanf("%d%d", &x, &y);
        int res = abs(x - y);
        res = min(res, d[0][x] + d[0][y]);
        res = min(res, d[1][x] + d[1][y]);
        res = min(res, d[0][x] + d[1][y] + dab);
        res = min(res, d[1][x] + d[0][y] + dab);
        printf("%d\n", res);
    }
    return 0;
}
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