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