cp/codeforces/920/e.cc
2025-09-09 09:47:57 +02:00

181 lines
3.5 KiB
C++

#include <bits/stdc++.h> // {{{
#include <version>
#ifdef __cpp_lib_ranges_enumerate
#include <ranges>
namespace rv = std::views;
namespace rs = std::ranges;
#endif
// https://codeforces.com/blog/entry/96344
#pragma GCC optimize("O2,unroll-loops")
#pragma GCC target("avx2,bmi,bmi2,lzcnt,popcnt")
using namespace std;
using i32 = int32_t;
using u32 = uint32_t;
using i64 = int64_t;
using u64 = uint64_t;
using f64 = double;
using f128 = long double;
#if __cplusplus >= 202002L
template <typename T>
constexpr T MIN = std::numeric_limits<T>::min();
template <typename T>
constexpr T MAX = std::numeric_limits<T>::max();
template <typename T>
[[nodiscard]] static T sc(auto&& x) {
return static_cast<T>(x);
}
template <typename T>
[[nodiscard]] static T sz(auto&& x) {
return static_cast<T>(x.size());
}
#endif
static void NO() {
std::cout << "NO\n";
}
static void YES() {
std::cout << "YES\n";
}
template <typename T>
using vec = std::vector<T>;
#define all(x) (x).begin(), (x).end()
#define rall(x) (x).rbegin(), (x).rend()
#define ff first
#define ss second
#ifdef LOCAL
#define db(...) std::print(__VA_ARGS__)
#define dbln(...) std::println(__VA_ARGS__)
#else
#define db(...)
#define dbln(...)
#endif
// }}}
void solve() {
i64 n, m;
cin >> n >> m;
i64 ar, ac, br, bc;
cin >> ar >> ac >> br >> bc;
// NOTE: forgot to adjust frontier + 1 (rly think thru!)
string ans{"Draw"};
if (ar < br) {
i64 frontier = (u64)ceill((ar + br) / 2.0);
// NOTE: key is alice gets extra move
// TBH, still confused (see mistaken note above) - don't understand
// didn't think about "optimal" play - i.e. alice leading one way
// can bob just go the opposite? then, alice ought to lead straight
// i think if u dig into the logic alice can lead straight in which bob
// can't do antyhing, but anyway all of this logic circumvented my mind, smh
// NOTE: this is why - SIMPLIFY
// simply consider the two player's positions on the frontier row itself -
// both on the same alice goes first so she gets extra move if frontier
// distance odd also, optimal strategy principle (explore too)
i64 aleft = max((i64)1, ac - (frontier - ar));
i64 aright = min(m, ac + frontier - ar);
i64 bleft = max((i64)1, bc - (br - frontier));
i64 bright = min(m, bc + (br - frontier));
bool alice_aggressor = (br - ar) & 1;
if (alice_aggressor) {
if (aleft <= bleft && aright >= bright) {
ans = "Alice";
}
} else {
if (bleft <= aleft && bright >= aright) {
ans = "Bob";
}
}
}
println("{}", ans);
/*
--A-------
-A-A------
BB--------
B---------
----------
----------
----------
----------
----------
----------
Alice
Bob
Draw
Draw
Draw
Alice
Draw
Draw
Bob
Alice
Alice
Draw
idea: consider frontier
note: ar >= br -> DRAW
ow, alice below bob, and there is a frontier
alice/bob have frontier pos - enumerate them
find aggressor - they cannot lose
if aggressor can take all passive piece -> win, print aggressor
ow, draw
chess moment :sunglasses:
O(h) -> ok
to flesh out:
derive frontier row
derive a/b cols on row
derive aggressor
note: only care about the coords
*/
}
int main() { // {{{
std::cin.exceptions(std::cin.failbit);
#ifdef LOCAL
std::cerr.rdbuf(std::cout.rdbuf());
std::cout.setf(std::ios::unitbuf);
std::cerr.setf(std::ios::unitbuf);
#else
std::cin.tie(nullptr)->sync_with_stdio(false);
#endif
u32 tc = 1;
std::cin >> tc;
for (u32 t = 0; t < tc; ++t) {
solve();
}
return 0;
}
// }}}