POJ1101(The Game)
The Game
| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 3058 | Accepted: 899 |
Description
One morning, you wake up and think: "I am such a good programmer. Why not make some money?'' So you decide to write a computer game.
The game takes place on a rectangular board consisting of w * h squares. Each square might or might not contain a game piece, as shown in the picture.
One important aspect of the game is whether two game pieces can be connected by a path which satisfies the two following properties:
It consists of straight segments, each one being either horizontal or vertical.
It does not cross any other game pieces.
(It is allowed that the path leaves the board temporarily.)
Here is an example:
The game pieces at (1,3) and at (4, 4) can be connected. The game pieces at (2, 3) and (3, 4) cannot be connected; each path would cross at least one other game piece.
The part of the game you have to write now is the one testing whether two game pieces can be connected according to the rules above.
The game takes place on a rectangular board consisting of w * h squares. Each square might or might not contain a game piece, as shown in the picture.
One important aspect of the game is whether two game pieces can be connected by a path which satisfies the two following properties:
It consists of straight segments, each one being either horizontal or vertical.
It does not cross any other game pieces.
(It is allowed that the path leaves the board temporarily.)
Here is an example:
The game pieces at (1,3) and at (4, 4) can be connected. The game pieces at (2, 3) and (3, 4) cannot be connected; each path would cross at least one other game piece.
The part of the game you have to write now is the one testing whether two game pieces can be connected according to the rules above.
Input
The input contains descriptions of several different game situations. The first line of each description contains two integers w and h (1 <= w,h <= 75), the width and the height of the board. The next h lines describe the contents of the board; each of these lines contains exactly w characters: a "X" if there is a game piece at this location, and a space if there is no game piece.
Each description is followed by several lines containing four integers x1, y1, x2, y2 each satisfying 1 <= x1,x2 <= w, 1 <= y1,y2 <= h. These are the coordinates of two game pieces. (The upper left corner has the coordinates (1, 1).) These two game pieces will always be different. The list of pairs of game pieces for a board will be terminated by a line containing "0 0 0 0".
The entire input is terminated by a test case starting with w=h=0. This test case should not be procesed.
Each description is followed by several lines containing four integers x1, y1, x2, y2 each satisfying 1 <= x1,x2 <= w, 1 <= y1,y2 <= h. These are the coordinates of two game pieces. (The upper left corner has the coordinates (1, 1).) These two game pieces will always be different. The list of pairs of game pieces for a board will be terminated by a line containing "0 0 0 0".
The entire input is terminated by a test case starting with w=h=0. This test case should not be procesed.
Output
For each board, output the line "Board #n:", where n is the number of the board. Then, output one line for each pair of game pieces associated with the board description. Each of these lines has to start with "Pair m: ", where m is the number of the pair (starting the count with 1 for each board). Follow this by "ksegments.", where k is the minimum number of segments for a path connecting the two game pieces, or "impossible.", if it is not possible to connect the two game pieces as described above.
Output a blank line after each board.
Output a blank line after each board.
Sample Input
5 4 XXXXX X X XXX X XXX 2 3 5 3 1 3 4 4 2 3 3 4 0 0 0 0 0 0
Sample Output
Board #1: Pair 1: 4 segments. Pair 2: 3 segments. Pair 3: impossible.
1
/*
2
It consists of straight segments, each one being either horizontal or vertical.
3
4
5
It does not cross any other game pieces.*/
6
#include <iostream>
7#include <queue>
8#define NIL INT_MAX
9using namespace std;
10
11const int N = 80;
12
13struct Point
14{
15 int x,y;
16 int cnt;//记录转化方向次数
17 bool operator <(const Point &tt) const
18
{
19 return cnt > tt.cnt;
20
}
21};
22
23int graph[N][N];
24int Visited[N][N];//记录转向次数
25int turn[N][N];//记录转向方向
26int mk[N][N];
27int w, h;
28int m, n;
29int dir[4][2] = {{0, 1}, {0, -1}, {1, 0}, {-1, 0}};
30
31bool isBound(int x, int y)
32{
33 if(x < 0 || y < 0)
34 return false;
35 if(x > m || y > n)
36 return false;
37 return true;
38}
39
40int bfs(Point bp, Point ep)
41{
42 Point a, b;
43 int i, j;
44 priority_queue<Point> Q;
45
46 //<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<
47
48 for(i = 0; i <= m; i++)
49 for(j = 0; j <= n; j++)
50 {
51 Visited[i][j] = NIL;
52 mk[i][j] = 0;
53 }
54
55 Visited[bp.x][bp.y] = 0;
56 turn[bp.x][bp.y] = -1;
57 mk[bp.x][bp.y] = 1;
58
59 for(i = 0; i < 4; i++)
60 {
61 a.x = bp.x + dir[i][0];
62 a.y = bp.y + dir[i][1];
63 a.cnt = 1;
64
65 Visited[a.x][a.y] = a.cnt;//记录拐弯次数
66 turn[a.x][a.y] = i;//记录拐弯方向
67
68 Q.push(a);
69 }
70
71 //<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<
72
73 while(!Q.empty())
74 {
75 a = Q.top();
76 Q.pop();
77
78 if(a.x == ep.x && a.y == ep.y)
79 return Visited[a.x][a.y];
80 if(graph[a.x][a.y])
81 continue;
82
83 mk[i][j] = 1;
84 for(i = 0; i < 4; i++)
85 {
86 b.x = a.x + dir[i][0];
87 b.y = a.y + dir[i][1];
88
89 if(isBound(b.x, b.y) && !mk[b.x][b.y])
90 {
91 if(i == turn[a.x][a.y])
92 b.cnt = Visited[a.x][a.y];
93 else
94 b.cnt = Visited[a.x][a.y] + 1;
95
96 if(b.cnt < Visited[b.x][b.y])
97 {
98 Visited[b.x][b.y] = b.cnt;
99 turn[b.x][b.y] = i;
100 Q.push(b);
101 }
102
103 }//for
104 }
105 }//while
106 return -1;
107}
108
109int main()
110{
111 char tc;
112 int i, j, ans, cc = 1, kk;
113 Point bp, ep;
114 while(scanf("%d%d", &w, &h))
115 {
116 //cc = 1;
117 m = h + 1;//m行
118 n = w + 1;//n列
119
120 if(!w && !h)
121 break;//输入结束
122
123 for(i = 0; i <= m; i++)
124 for(j = 0; j <= n; j++)
125 graph[i][j] = 0;//初始化
126 for(i = 1; i <= h; i++)
127 {
128 getchar();
129 for(j = 1; j <= w; j++)
130 {
131 scanf("%c", &tc);
132 if(tc == 'X')
133 graph[i][j] = 1;
134 }
135 }
136 cout<<"Board #"<<cc++<<":"<<endl;
137 kk = 1;
138 while(scanf("%d%d%d%d", &bp.y, &bp.x, &ep.y, &ep.x))
139 {
140 if(!bp.x && !bp.y && !ep.x && !ep.y)
141 break;//测试用例输入结束
142 ans = bfs(bp, ep);
143 if(ans != -1)
144 cout<<"Pair "<<kk++<<": "<<ans<<" segments."<<endl;
145 else
146 cout<<"Pair "<<kk++<<": impossible."<<endl;
147 }
148 cout<<endl;
149 }
150 return 0;
151}
152
/*
2
It consists of straight segments, each one being either horizontal or vertical. 3
4
5
It does not cross any other game pieces.*/6
#include <iostream>7
#include <queue>8
#define NIL INT_MAX9
using namespace std;10
11
const int N = 80;12
13
struct Point14
{
15
int x,y;16
int cnt;//记录转化方向次数17
bool operator <(const Point &tt) const18
{
19
return cnt > tt.cnt;20
}21
};22
23
int graph[N][N];24
int Visited[N][N];//记录转向次数25
int turn[N][N];//记录转向方向26
int mk[N][N];27
int w, h;28
int m, n;29
int dir[4][2] = {{0, 1}, {0, -1}, {1, 0}, {-1, 0}};30
31
bool isBound(int x, int y)32
{33
if(x < 0 || y < 0)34
return false;35
if(x > m || y > n)36
return false;37
return true;38
}39
40
int bfs(Point bp, Point ep)41
{42
Point a, b;43
int i, j;44
priority_queue<Point> Q;45
46
//<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<47
48
for(i = 0; i <= m; i++)49
for(j = 0; j <= n; j++)50
{51
Visited[i][j] = NIL;52
mk[i][j] = 0;53
}54
55
Visited[bp.x][bp.y] = 0;56
turn[bp.x][bp.y] = -1;57
mk[bp.x][bp.y] = 1;58
59
for(i = 0; i < 4; i++)60
{61
a.x = bp.x + dir[i][0];62
a.y = bp.y + dir[i][1];63
a.cnt = 1;64
65
Visited[a.x][a.y] = a.cnt;//记录拐弯次数66
turn[a.x][a.y] = i;//记录拐弯方向67
68
Q.push(a);69
}70
71
//<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<72
73
while(!Q.empty())74
{75
a = Q.top();76
Q.pop();77
78
if(a.x == ep.x && a.y == ep.y)79
return Visited[a.x][a.y];80
if(graph[a.x][a.y])81
continue;82
83
mk[i][j] = 1;84
for(i = 0; i < 4; i++)85
{86
b.x = a.x + dir[i][0];87
b.y = a.y + dir[i][1];88
89
if(isBound(b.x, b.y) && !mk[b.x][b.y])90
{ 91
if(i == turn[a.x][a.y])92
b.cnt = Visited[a.x][a.y];93
else 94
b.cnt = Visited[a.x][a.y] + 1;95
96
if(b.cnt < Visited[b.x][b.y])97
{98
Visited[b.x][b.y] = b.cnt;99
turn[b.x][b.y] = i;100
Q.push(b);101
}102
103
}//for104
}105
}//while106
return -1;107
}108
109
int main()110
{111
char tc;112
int i, j, ans, cc = 1, kk;113
Point bp, ep;114
while(scanf("%d%d", &w, &h))115
{116
//cc = 1;117
m = h + 1;//m行118
n = w + 1;//n列119
120
if(!w && !h)121
break;//输入结束122
123
for(i = 0; i <= m; i++)124
for(j = 0; j <= n; j++)125
graph[i][j] = 0;//初始化126
for(i = 1; i <= h; i++)127
{128
getchar();129
for(j = 1; j <= w; j++)130
{131
scanf("%c", &tc);132
if(tc == 'X')133
graph[i][j] = 1;134
}135
}136
cout<<"Board #"<<cc++<<":"<<endl;137
kk = 1;138
while(scanf("%d%d%d%d", &bp.y, &bp.x, &ep.y, &ep.x))139
{140
if(!bp.x && !bp.y && !ep.x && !ep.y)141
break;//测试用例输入结束142
ans = bfs(bp, ep);143
if(ans != -1)144
cout<<"Pair "<<kk++<<": "<<ans<<" segments."<<endl;145
else146
cout<<"Pair "<<kk++<<": impossible."<<endl;147
}148
cout<<endl;149
}150
return 0;151
}152
