/**
* 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;
}
