HDU-2612-Find a way

Posted 2020-12-24 1625--h

Find a way

Problem 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.
‘@’ KCF

 

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

 

4 4

Y.#@

....

.#..

@..M

4 4

Y.#@

....

.#..

@#.M

5 5

[email protected]

.#...

.#...

@..M.

#...#

 

Sample Output

66

88

66

 

分析

1. 需要单独记录下两人的坐标，我们可以在输入的时候就可以记录下来。

2. 要求他两到同一个KFC的最短时间，所以再广搜的时候，要在队列元素中有一个记录时间的元素。

细节

要保证KFC两个人都可以到，才可以进行比较

 

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 
 4 struct node
 5 {
 6     int x, y, t;
 7     node(int a, int b, int c)
 8     {
 9         x = a, y = b, t = c;
10     }
11 };
12 
13 char mp[203][203];
14 int arr[203][203], may[203][203];
15 bool mk[203][203];
16 int dir[4][2] = { {-1, 0}, {1, 0}, {0, -1}, {0, 1} };
17 
18 void bfs(int bgx, int bgy)
19 {
20     memset(mk, false, sizeof(mk));
21     queue<node> que;
22     que.push(node(bgx, bgy, 0));
23     mk[bgx][bgy] = true;
24     while( !que.empty())
25     {
26         node f = que.front();
27         que.pop();
28         for(int i=0; i<4; ++ i)
29         {
30             int x = f.x + dir[i][0], y = f.y + dir[i][1];
31             int t = f.t + 1;
32             if(mp[x][y] != ‘#‘ && mk[x][y] == false)
33             {
34                 mk[x][y] = true;
35                 if(mp[x][y] == ‘@‘)
36                     arr[x][y] += t, may[x][y] += 1;
37                 que.push(node(x, y, t));
38             }
39         }
40     }
41 }
42 int main()
43 {
44     int n, m;
45     while(scanf("%d%d", &n, &m) != EOF)
46     {
47         int y_bgx, y_bgy, m_bgx, m_bgy;
48         memset(mp, ‘#‘, sizeof(mp));
49         for(int i=1; i<=n; ++ i)
50         {
51             for(int j=1; j<=m; ++ j)
52             {
53                 scanf(" %c", &mp[i][j]);
54                 if(mp[i][j] == ‘Y‘)
55                     y_bgx = i, y_bgy = j;
56                 if(mp[i][j] == ‘M‘)
57                     m_bgx = i, m_bgy = j;
58             }
59         }
60         memset(arr, 0, sizeof(arr));
61         memset(may, 0, sizeof(may));
62 
63         bfs(y_bgx, y_bgy);
64         bfs(m_bgx, m_bgy);
65         int ans = 1000000;
66         for(int i=1; i<=n; ++ i)
67         {
68             for(int j=1; j<=m; ++ j)
69             {
70                 if(mp[i][j] == ‘@‘ && may[i][j] == 2)
71                 {
72                     if(arr[i][j] < ans)
73                         ans = arr[i][j];
74                 }
75             }
76         }
77         printf("%d
", ans * 11);
78     }
79     return 0;
80 }