forked from gouthampradhan/leetcode
-
Notifications
You must be signed in to change notification settings - Fork 0
/
CatAndMouse.java
152 lines (140 loc) · 4.95 KB
/
CatAndMouse.java
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
121
122
123
124
125
126
127
128
129
130
131
132
133
134
135
136
137
138
139
140
141
142
143
144
145
146
147
148
149
150
151
152
package dynamic_programming;
import java.util.*;
/**
* Created by gouthamvidyapradhan on 03/03/2019
*
* <p>A game on an undirected graph is played by two players, Mouse and Cat, who alternate turns.
*
* <p>The graph is given as follows: graph[a] is a list of all nodes b such that ab is an edge of
* the graph.
*
* <p>Mouse starts at node 1 and goes first, Cat starts at node 2 and goes second, and there is a
* Hole at node 0.
*
* <p>During each player's turn, they must travel along one edge of the graph that meets where they
* are. For example, if the Mouse is at node 1, it must travel to any node in graph[1].
*
* <p>Additionally, it is not allowed for the Cat to travel to the Hole (node 0.)
*
* <p>Then, the game can end in 3 ways:
*
* <p>If ever the Cat occupies the same node as the Mouse, the Cat wins. If ever the Mouse reaches
* the Hole, the Mouse wins. If ever a position is repeated (ie. the players are in the same
* position as a previous turn, and it is the same player's turn to move), the game is a draw. Given
* a graph, and assuming both players play optimally, return 1 if the game is won by Mouse, 2 if the
* game is won by Cat, and 0 if the game is a draw.
*
* <p>Example 1:
*
* <p>Input: [[2,5],[3],[0,4,5],[1,4,5],[2,3],[0,2,3]] Output: 0 Explanation: 4---3---1 | | 2---5 \
* / 0
*
* <p>Note:
*
* <p>3 <= graph.length <= 50 It is guaranteed that graph[1] is non-empty. It is guaranteed that
* graph[2] contains a non-zero element.
*
* <p>Solution: Each combination of mouse, cat and turn for a state (m, c, p). Transition between
* different states form a state graph - start to color each state with either MOUSE, CAT or a DRAW
* based on who wins this state. Perform a bottom up dp inorder to arrive at the answer.
*/
public class CatAndMouse {
private final int CAT = 2, MOUSE = 1, DRAW = 0, PLAYER_CAT = 1, PLAYER_MOUSE = 0;
private Queue<State> queue = new ArrayDeque<>();
class State {
int catPos, mPos, player, color;
State(int mPos, int catPos, int player, int color) {
this.catPos = catPos;
this.mPos = mPos;
this.player = player;
this.color = color;
}
}
/**
* Main method
*
* @param args
*/
public static void main(String[] args) {
int[][] graph = {{2, 5}, {3}, {0, 4, 5}, {1, 4, 5}, {2, 3}, {0, 2, 3}};
System.out.println(new CatAndMouse().catMouseGame(graph));
}
public int catMouseGame(int[][] graph) {
int[][][] color = new int[graph.length][graph.length][2];
int[][][] degree = new int[graph.length][graph.length][2];
for (int i = 1; i < graph.length; i++) {
for (int p = 0; p < 2; p++) {
color[0][i][p] = MOUSE;
queue.offer(new State(0, i, p, MOUSE));
color[i][i][p] = CAT;
queue.offer(new State(i, i, p, CAT));
}
}
for (int m = 0; m < graph.length; m++) {
for (int c = 1; c < graph.length; c++) {
degree[m][c][0] = graph[m].length;
degree[m][c][1] = graph[c].length;
for (int v : graph[c]) {
if (v == 0) {
degree[m][c][1]--;
break;
}
}
}
}
while (!queue.isEmpty()) {
State current = queue.poll();
List<State> parents = getParents(graph, current, color);
if (color[current.mPos][current.catPos][current.player] == CAT) {
enqueue(queue, parents, PLAYER_CAT, CAT, color, degree);
} else {
enqueue(queue, parents, PLAYER_MOUSE, MOUSE, color, degree);
}
}
return color[1][2][0];
}
private void enqueue(
Queue<State> queue,
List<State> parents,
int player,
int col,
int[][][] color,
int[][][] degree) {
for (State parent : parents) {
if (color[parent.mPos][parent.catPos][parent.player] == DRAW) {
if (parent.player == player) {
color[parent.mPos][parent.catPos][parent.player] = col;
queue.offer(new State(parent.mPos, parent.catPos, parent.player, col));
} else {
int currDegree = --degree[parent.mPos][parent.catPos][parent.player];
if (currDegree == 0) {
color[parent.mPos][parent.catPos][parent.player] = col;
queue.offer(new State(parent.mPos, parent.catPos, parent.player, col));
}
}
}
}
}
private List<State> getParents(int[][] graph, State current, int[][][] color) {
int player = current.player;
int[] positions;
List<State> list = new ArrayList<>();
if (player == PLAYER_MOUSE) {
positions = graph[current.catPos];
for (int pos : positions) {
if (pos == 0) continue;
if (color[current.mPos][pos][PLAYER_CAT] == DRAW) {
list.add(new State(current.mPos, pos, PLAYER_CAT, DRAW));
}
}
} else {
positions = graph[current.mPos];
for (int pos : positions) {
if (color[pos][current.catPos][PLAYER_MOUSE] == DRAW) {
list.add(new State(pos, current.catPos, PLAYER_MOUSE, DRAW));
}
}
}
return list;
}
}