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ID | 题目 | 提交者 | 结果 | 用时 | 内存 | 语言 | 文件大小 | 提交时间 | 测评时间 |
---|---|---|---|---|---|---|---|---|---|
#285334 | #6633. Exam Requirements | Misuki | ML | 1ms | 3516kb | C++20 | 7.0kb | 2023-12-16 17:57:50 | 2023-12-16 17:57:51 |
Judging History
answer
#pragma GCC optimize("O2")
#include <algorithm>
#include <array>
#include <bit>
#include <bitset>
#include <cassert>
#include <cctype>
#include <cfenv>
#include <cfloat>
#include <chrono>
#include <cinttypes>
#include <climits>
#include <cmath>
#include <compare>
#include <complex>
#include <concepts>
#include <cstdarg>
#include <cstddef>
#include <cstdint>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <deque>
#include <fstream>
#include <functional>
#include <initializer_list>
#include <iomanip>
#include <ios>
#include <iostream>
#include <istream>
#include <iterator>
#include <limits>
#include <list>
#include <map>
#include <memory>
#include <new>
#include <numbers>
#include <numeric>
#include <ostream>
#include <queue>
#include <random>
#include <ranges>
#include <set>
#include <span>
#include <sstream>
#include <stack>
#include <streambuf>
#include <string>
#include <tuple>
#include <type_traits>
#include <variant>
//#define int ll
#define INT128_MAX (__int128)(((unsigned __int128) 1 << ((sizeof(__int128) * __CHAR_BIT__) - 1)) - 1)
#define INT128_MIN (-INT128_MAX - 1)
namespace R = std::ranges;
namespace V = std::views;
using namespace std;
using ll = long long;
using ull = unsigned long long;
using pii = pair<int, int>;
using pll = pair<long long, long long>;
using tiii = tuple<int, int, int>;
using ldb = long double;
//#define double ldb
template<class T>
ostream& operator<<(ostream& os, const pair<T, T> pr) {
return os << pr.first << ' ' << pr.second;
}
template<class T, size_t N>
ostream& operator<<(ostream& os, const array<T, N> &arr) {
for(const T &X : arr)
os << X << ' ';
return os;
}
template<class T>
ostream& operator<<(ostream& os, const vector<T> &vec) {
for(const T &X : vec)
os << X << ' ';
return os;
}
#define rep(i, a, b) for(int i = a; i < (b); ++i)
#define all(x) begin(x), end(x)
#define sz(x) (int)(x).size()
typedef vector<int> vi;
/**
* Author: Emil Lenngren, Simon Lindholm
* Date: 2011-11-29
* License: CC0
* Source: folklore
* Description: Calculates a valid assignment to boolean variables a, b, c,... to a 2-SAT problem, so that an expression of the type $(a\|\|b)\&\&(!a\|\|c)\&\&(d\|\|!b)\&\&...$ becomes true, or reports that it is unsatisfiable.
* Negated variables are represented by bit-inversions (\texttt{\tilde{}x}).
* Usage:
* TwoSat ts(number of boolean variables);
* ts.either(0, \tilde3); // Var 0 is true or var 3 is false
* ts.setValue(2); // Var 2 is true
* ts.atMostOne({0,\tilde1,2}); // <= 1 of vars 0, \tilde1 and 2 are true
* ts.solve(); // Returns true iff it is solvable
* ts.values[0..N-1] holds the assigned values to the vars
* Time: O(N+E), where N is the number of boolean variables, and E is the number of clauses.
* Status: stress-tested
*/
#define rep(i, a, b) for(int i = a; i < (b); ++i)
#define sz(x) (int)(x).size()
struct TwoSat {
int N;
vector<vi> gr;
vi values; // 0 = false, 1 = true
TwoSat(int n = 0) : N(n), gr(2*n) {}
int addVar() { // (optional)
//gr.emplace_back();
//gr.emplace_back();
return N++;
}
void either(int f, int j) {
f = max(2*f, -1-2*f);
j = max(2*j, -1-2*j);
gr[f].push_back(j^1);
gr[j].push_back(f^1);
}
void setValue(int x) { either(x, x); }
void atMostOne(const vi& li) { // (optional)
if (sz(li) <= 1) return;
int cur = ~li[0];
rep(i,2,sz(li)) {
int next = addVar();
either(cur, ~li[i]);
either(cur, next);
either(~li[i], next);
cur = ~next;
}
either(cur, ~li[1]);
}
vi val, comp, z; int time = 0;
int dfs(int i) {
int low = val[i] = ++time, x; z.push_back(i);
for(int e : gr[i]) if (!comp[e])
low = min(low, val[e] ?: dfs(e));
if (low == val[i]) do {
x = z.back(); z.pop_back();
comp[x] = low;
if (values[x>>1] == -1)
values[x>>1] = x&1;
} while (x != i);
return val[i] = low;
}
bool solve() {
values.assign(N, -1);
val.assign(2*N, 0); comp = val;
rep(i,0,2*N) if (!comp[i]) dfs(i);
rep(i,0,N) if (comp[2*i] == comp[2*i+1]) return 0;
return 1;
}
};
signed main() {
ios::sync_with_stdio(false), cin.tie(NULL);
int t; cin >> t;
while(t--) {
int n, m; cin >> n >> m;
vector<array<int, 2>> lr(n), cond(m);
for(auto &[l, r] : lr)
cin >> l >> r;
for(auto &[x, y] : cond) {
cin >> x >> y;
x--, y--;
}
{
vector<int> tmp;
for(auto [l, r] : lr) {
tmp.emplace_back(l);
tmp.emplace_back(r);
}
R::sort(tmp);
tmp.resize(unique(tmp.begin(), tmp.end()) - tmp.begin());
for(auto &[l, r] : lr) {
l = R::lower_bound(tmp, l) - tmp.begin();
r = R::lower_bound(tmp, r) - tmp.begin();
}
}
const int C = bit_ceil((unsigned)(n << 1));
vector<vector<int>> seg(C << 1);
auto query = [&](int l, int r) {
vector<int> res;
for(l += C, r += C; l < r; l >>= 1, r >>= 1) {
if (l & 1) res.emplace_back(l++);
if (r & 1) res.emplace_back(--r);
}
return res;
};
int fakeN = n + 2 * C;
vector<int> ssz(2 * fakeN);
{ //precalc size
auto fakeAddVar = [&]() {
ssz.emplace_back();
ssz.emplace_back();
return fakeN++;
};
auto fakeEither = [&](int f, int j) {
f = max(2*f, -1-2*f);
j = max(2*j, -1-2*j);
ssz[f]++;
ssz[j]++;
};
auto fakeAtMostOne = [&](const vi& li) {
if (sz(li) <= 1) return;
int cur = ~li[0];
rep(i,2,sz(li)) {
int next = fakeAddVar();
fakeEither(cur, ~li[i]);
fakeEither(cur, next);
fakeEither(~li[i], next);
cur = ~next;
}
fakeEither(cur, ~li[1]);
};
for(auto [x, y] : cond)
fakeEither(x, y);
for(int i = C; i < 2 * C; i++) {
vector<int> path;
int now = i;
while(now) {
path.emplace_back(now + n);
now >>= 1;
}
fakeAtMostOne(path);
}
for(int i = 0; auto [l, r] : lr) {
for(int x : query(l, r + 1))
seg[x].emplace_back(i);
i++;
}
for(int i = 1; i < 2 * C; i++) {
if (!seg[i].empty())
fakeAtMostOne(seg[i]);
for(int x : seg[i])
fakeEither(~x, i + n);
}
}
TwoSat ts(n + 2 * C);
ts.gr.resize(ssize(ssz));
for(int i = 0; i < ssize(ssz); i++)
ts.gr[i].reserve(ssz[i]);
for(auto [x, y] : cond)
ts.either(x, y);
for(int i = C; i < 2 * C; i++) {
vector<int> path;
int now = i;
while(now) {
path.emplace_back(now + n);
now >>= 1;
}
ts.atMostOne(path);
}
for(int i = 1; i < 2 * C; i++) {
if (!seg[i].empty())
ts.atMostOne(seg[i]);
for(int x : seg[i])
ts.either(~x, i + n);
}
cout << (ts.solve() ? "YES\n" : "NO\n");
}
return 0;
}
詳細信息
Test #1:
score: 100
Accepted
time: 1ms
memory: 3516kb
input:
2 3 1 1 5 2 7 10 11 2 1 3 3 1 5 2 7 5 7 1 2 2 3 3 1
output:
YES NO
result:
ok 2 lines
Test #2:
score: -100
Memory Limit Exceeded
input:
1 100000 100000 15084647 15220430 217541056 217598054 222963701 223110976 71224450 71340221 320463268 320631387 334861459 334924668 328950591 329083669 178996498 178996584 428529461 428605033 405428435 405504132 197936687 197947412 9058562 9190197 169407597 169474101 297518153 297590927 31471874 316...
output:
NO