Hdu2612 Find a way

发表于 | 更新于: | 分类于 ， ， | | 阅读数
字数统计: 951 | 阅读时长 ≈ 5

Description

Pass a year learning in Hangzhou, yifenfei arrival hometown Ningbo at finally. Leave Ningbo one year, yifenfei have many people to meet. Especially a good friend Merceki.
Yifenfei’s home is at the countryside, but Merceki’s home is in the center of city. So yifenfei made arrangements with Merceki to meet at a KFC. There are many KFC in Ningbo, they want to choose one that let the total time to it be most smallest.
Now give you a Ningbo map, Both yifenfei and Merceki can move up, down ,left, right to the adjacent road by cost 11 minutes.

Input

The input contains multiple test cases.
Each test case include, first two integers n, m. (2<=n,m<=200).
Next n lines, each line included m character.
‘Y’ express yifenfei initial position.
‘M’ express Merceki initial position.
‘#’ forbid road;
‘.’ Road.
‘@’ KFC

Output

For each test case output the minimum total time that both yifenfei and Merceki to arrival one of KFC.You may sure there is always have a KFC that can let them meet.

Sample Input

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
4 4
Y.#@
....
.#..
@..M
4 4
Y.#@
....
.#..
@#.M
5 5
Y..@.
.#...
.#...
@..M.
#...#

Sample Output

1
2
3
66
88
66

Analysis

求一点到彼此所用时间之和最小，可以将y和m到每点的时间分别用BFS打表存储，然后扫到’@‘后计算，算出最小的一个时间。另外y和m不能经过互相的出发点，所以在他们分别BFS时，地图上对方需暂时变为‘#’。

Code

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
#include <iostream>
#include <queue>
#include <cstring>
#include <algorithm>
#include <cstdio>
int dby[1000][1000];
int dbm[1000][1000];
bool visit[1000][1000];
char map[1000][1000];
struct  node
{
	int n, m, t;
};
using namespace std;
int main()
{
	int n, m;
	while (cin >> n >> m)
	{
		int ans = 1000000;
		int yn, ym, mn, mm;
		memset(dby, 0, sizeof(dby));
		memset(dbm, 0, sizeof(dbm));
		for (int i = 0; i < n; i++)
		{
			cin >> map[i];
			for (int j = 0; j < m; j++)
			{
				if (map[i][j] == 'Y')
				{
					yn = i;
					ym = j;
				}
				if (map[i][j] == 'M')
				{
					mn = i;
					mm = j;
				}
			}
		}
		memset(visit, 0, sizeof(visit));
		node start;
		start.n = yn;
		start.m = ym;
		start.t = 0;
		visit[yn][ym] = 1;
		map[mn][mm] = '#';
		queue<node> qy;
		qy.push(start);
		while (!qy.empty())
		{
			node fy = qy.front();
			qy.pop();
			dby[fy.n][fy.m] = fy.t;
			fy.t += 1;
			node vy = fy;
			if (vy.n - 1 >= 0 && visit[vy.n - 1][vy.m] == 0 && map[vy.n - 1][vy.m] != '#')
			{
				visit[vy.n - 1][vy.m] = 1;
				dby[vy.n - 1][vy.m] = vy.t;
				vy.n -= 1;
				qy.push(vy);
			}
			vy = fy;
			if (vy.n + 1 < n && visit[vy.n + 1][vy.m] == 0 && map[vy.n + 1][vy.m] != '#')
			{
				visit[vy.n + 1][vy.m] = 1;
				dby[vy.n + 1][vy.m] = vy.t;
				vy.n += 1;
				qy.push(vy);
			}
			vy = fy;
			if (vy.m - 1 >= 0 && visit[vy.n][vy.m - 1] == 0 && map[vy.n][vy.m - 1] != '#')
			{
				visit[vy.n][vy.m - 1] = 1;
				dby[vy.n][vy.m - 1] = vy.t;
				vy.m -= 1;
				qy.push(vy);
			}
			vy = fy;
			if (vy.m + 1 < m && visit[vy.n][vy.m + 1] == 0 && map[vy.n][vy.m + 1] != '#')
			{
				visit[vy.n][vy.m + 1] = 1;
				dby[vy.n][vy.m + 1] = vy.t;
				vy.m += 1;
				qy.push(vy);
			}
		}
		map[mn][mm] = 'M';
		memset(visit, 0, sizeof(visit));
		start.n = mn;
		start.m = mm;
		start.t = 0;
		visit[mn][mm] = 1;
		map[yn][ym] = '#';
		queue<node> qm;
		qm.push(start);
		while (!qm.empty())
		{
			node fm = qm.front();
			qm.pop();
			dbm[fm.n][fm.m] = fm.t;
			fm.t += 1;
			node vm = fm;
			if (vm.n - 1 >= 0 && visit[vm.n - 1][vm.m] == 0 && map[vm.n - 1][vm.m] != '#')
			{
				visit[vm.n - 1][vm.m] = 1;
				dbm[vm.n - 1][vm.m] = vm.t;
				vm.n -= 1;
				qm.push(vm);
			}
			vm = fm;
			if (vm.n + 1 < n && visit[vm.n + 1][vm.m] == 0 && map[vm.n + 1][vm.m] != '#')
			{
				visit[vm.n + 1][vm.m] = 1;
				dbm[vm.n + 1][vm.m] = vm.t;
				vm.n += 1;
				qm.push(vm);
			}
			vm = fm;
			if (vm.m - 1 >= 0 && visit[vm.n][vm.m - 1] == 0 && map[vm.n][vm.m - 1] != '#')
			{
				visit[vm.n][vm.m - 1] = 1;
				dbm[vm.n][vm.m - 1] = vm.t;
				vm.m -= 1;
				qm.push(vm);
			}
			vm = fm;
			if (vm.m + 1 < m && visit[vm.n][vm.m + 1] == 0 && map[vm.n][vm.m + 1] != '#')
			{
				visit[vm.n][vm.m + 1] = 1;
				dbm[vm.n][vm.m + 1] = vm.t;
				vm.m += 1;
				qm.push(vm);
			}
		}
		map[yn][ym] = 'Y';
		for (int i = 0; i < n; i++)
		{
			for (int j = 0; j < m; j++)
			{
				if (map[i][j] == '@'&&dby[i][j] != 0 && dbm[i][j] != 0)
				{
					ans = min(ans, dby[i][j] + dbm[i][j]);
				}
			}
		}
		cout << ans * 11 << endl;
	}
	return 0;
}