이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
#include <bits/stdc++.h>
#define nl "\n"
#define no "NO"
#define yes "YES"
#define fi first
#define se second
#define vec vector
#define task "main"
#define _mp make_pair
#define ii pair<int, int>
#define sz(x) (int)x.size()
#define all(x) x.begin(), x.end()
#define evoid(val) return void(std::cout << val)
#define FOR(i, a, b) for(int i = (a); i <= (b); ++i)
#define FOD(i, b, a) for(int i = (b); i >= (a); --i)
#define unq(x) sort(all(x)); x.resize(unique(all(x)) - x.begin())
using namespace std;
template<typename U, typename V> bool maxi(U &a, V b) {
if (a < b) { a = b; return 1; } return 0;
}
template<typename U, typename V> bool mini(U &a, V b) {
if (a > b or a == -1) { a = b; return 1; } return 0;
}
const int N = (int)2e5 + 9;
const int mod = (int)1e9 + 7;
void prepare(); void main_code();
int main() {
ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
if (fopen(task".inp", "r")) {
freopen(task".inp", "r", stdin);
freopen(task".out", "w", stdout);
}
const bool MULTITEST = 0; prepare();
int num_test = 1; if (MULTITEST) cin >> num_test;
while (num_test--) { main_code(); }
}
void prepare() {};
int n, k, m, q;
int a[N], b[N];
pair<int, int> Q[N];
namespace SUB1 {
const int N = 307;
int d[N][N];
void sol() {
memset(d, -1, sizeof d);
FOR(i, 1, m) {
if (a[i] < b[i]) {
FOR(j, a[i], min(a[i] + k - 1, b[i] - 1))
FOR(t, j + 1, b[i])
d[j][t] = 1;
} else {
FOD(j, a[i], max(a[i] - k + 1, b[i] + 1))
FOD(t, j - 1, b[i])
d[j][t] = 1;
}
}
FOR(k, 1, n)
FOR(i, 1, n) if (d[i][k] != -1)
FOR(j, 1, n) if (d[k][j] != -1) {
mini(d[i][j], d[i][k] + d[k][j]);
}
FOR(i, 1, q) cout << d[Q[i].fi][Q[i].se] << nl;
}
};
namespace SUB2 {
int l[N], r[N], d[N];
vector<int> insmax[N], revmax[N], insmin[N], revmin[N];
void sol() {
FOR(i, 1, n) l[i] = n + 1;
FOR(i, 1, m) {
if (a[i] < b[i]) {
insmax[a[i]].push_back(b[i]);
revmax[min(a[i] + 1 - 1, b[i] - 1)].push_back(b[i]);
} else {
insmin[a[i]].push_back(b[i]);
revmin[max(a[i] - k + 1, b[i] + 1)].push_back(b[i]);
}
}
multiset<int> v;
FOR(i, 1, n) {
for(int j : insmax[i]) {
v.insert(j);
}
if (sz(v)) r[i] = *v.rbegin();
for(int j : revmax[i]) {
v.erase(v.find(j));
}
}
FOD(i, n, 1) {
for(int j : insmin[i]) {
v.insert(j);
}
if (sz(v)) l[i] = *v.begin();
for(int j : revmin[i]) {
v.erase(v.find(j));
}
}
FOR(i, 1, q) {
queue<int> f;
int x = Q[i].fi, y = Q[i].se;
int L = x, R = x;
FOR(i, 1, n) d[i] = -1;
d[x] = 0;
f.push(x);
while (!f.empty()) {
int u = f.front(); f.pop();
if (u == y) break ;
if (l[u] < L) {
FOR(j, l[u], L - 1) {
d[j] = d[u] + 1;
f.push(j);
}
L = l[u];
}
if (r[u] > R) {
FOR(j, R + 1, r[u]) {
d[j] = d[u] + 1;
f.push(j);
}
R = r[u];
}
}
cout << d[y] << nl;
}
}
};
void main_code() {
cin >> n >> k >> m;
for(int i = 1; i <= m; ++i) {
cin >> a[i] >> b[i];
}
cin >> q;
for(int i = 1; i <= q; ++i) {
cin >> Q[i].fi >> Q[i].se;
}
if (n <= 300 and m <= 300) {
return SUB1::sol(), void();
}
if (max(n, max(m, q)) <= 2000 or q == 1) {
return SUB2::sol(), void();
}
}
/* Let the river flows naturally */
컴파일 시 표준 에러 (stderr) 메시지
Main.cpp: In function 'int main()':
Main.cpp:36:16: warning: ignoring return value of 'FILE* freopen(const char*, const char*, FILE*)' declared with attribute 'warn_unused_result' [-Wunused-result]
36 | freopen(task".inp", "r", stdin);
| ~~~~~~~^~~~~~~~~~~~~~~~~~~~~~~~
Main.cpp:37:16: warning: ignoring return value of 'FILE* freopen(const char*, const char*, FILE*)' declared with attribute 'warn_unused_result' [-Wunused-result]
37 | freopen(task".out", "w", stdout);
| ~~~~~~~^~~~~~~~~~~~~~~~~~~~~~~~~
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