Submission #398220

#TimeUsernameProblemLanguageResultExecution timeMemory
398220marsCop and Robber (BOI14_coprobber)C++14
100 / 100
548 ms7556 KiB
/** * Official solution for Cop and Robber. * * Implements the idea described in the spoiler. LeftToWin array * counts how many winning moves are necessary to declare the position * as winning. The cop's winning strategy is stored in NextPos array. * * This file must be compiled in C++11 mode. * * Author: Marijonas Petrauskas */ #include "coprobber.h" #include <algorithm> #include <queue> #include <tuple> using namespace std; typedef tuple<int, int, int> Position; const int COP = 0, ROBBER = 1; int LeftToWin[2][MAX_N][MAX_N]; int NextPos[MAX_N][MAX_N]; int start(int N, bool A[MAX_N][MAX_N]) { for (int r = 0; r < N; r++) { int degree = count(A[r], A[r] + N, true); for (int c = 0; c < N; c++) if (c != r) { LeftToWin[COP][c][r] = 1; LeftToWin[ROBBER][c][r] = degree; } } queue<Position> q; for (int i = 0; i < N; i++) { q.push(Position(COP, i, i)); q.push(Position(ROBBER, i, i)); } int numProcessed = 0; while (!q.empty()) { int t, c, r; tie(t, c, r) = q.front(); q.pop(); numProcessed++; if (t == COP) { for (int n = 0; n < N; n++) if (A[r][n] && LeftToWin[ROBBER][c][n]) { LeftToWin[ROBBER][c][n]--; if (LeftToWin[ROBBER][c][n] == 0) q.push(Position(ROBBER, c, n)); } } else if (t == ROBBER) { for (int n = 0; n < N; n++) if ((c == n || A[c][n]) && LeftToWin[COP][n][r]) { LeftToWin[COP][n][r] = 0; q.push(Position(COP, n, r)); NextPos[n][r] = c; } } } return numProcessed == 2 * N * N ? 0 : -1; } int cop = 0; int nextMove(int robber) { cop = NextPos[cop][robber]; return cop; }
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