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.
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.
InputThe 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
OutputFor 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
Y..@.
.#...
.#...
@..M.
#...#
Sample Output
66 88 66
题目意思:
有二个人Y和M,俩个人要到@(肯德基)初见面,求最短路径和;
解法:
也是一个使用BFS的题目
1 #include <iostream>
2 #include <cstring>
3 #include <queue>
4 using namespace std;
5
6 const int MAX = 200 + 199;
7 char Map[MAX][MAX];
8 int visit[2][MAX][MAX];
9 int dir[4][2] = { {-1,0},{1,0},{0,-1},{0,1} };
10 int x1,y1;
11 int x2,y2;
12 int L,W;
13
14 struct S
15 {
16 int x,y;
17 int cont;
18 };
19
20 void bfs(int x,int y,int ti)
21 {
22 queue<S>P;
23 S ss;
24 ss.x = x;
25 ss.y = y;
26 ss.cont = 0;
27 visit[ti][x][y] = 0;
28 P.push(ss);
29 //cout<<" ti === "<<ti<<endl;
30 while(!P.empty())
31 {
32 S t;
33 t = P.front();
34 P.pop();
35 //cout<<" ******** "<<t.x<<t.y<<endl;
36 for(int i = 0;i < 4;i++)
37 {
38 S temp ;
39 temp.x = t.x + dir[i][0];
40 temp.y = t.y + dir[i][1];
41 temp.cont = t.cont+1;
42
43 if(temp.x < 1||temp.x > L||temp.y < 1||temp.y > W)
44 continue;
45 //cout<<"temp.x = "<<temp.x<<"temp.y = "<<temp.y<<"temp.cont = "<<temp.cont<<endl;
46 if(Map[temp.x][temp.y] !='#'&& visit[ti][temp.x][temp.y] == -1)
47 {
48 visit[ti][temp.x][temp.y] = temp.cont;
49 P.push(temp);
50 }
51 }
52
53 }
54
55 }
56
57 int main()
58 {
59 while(cin>>L>>W)
60 {
61 memset(Map,'0',sizeof(Map));
62 memset(visit,-1,sizeof(visit));
63
64
65 for(int i =1;i <= L;i++)
66 for(int j = 1;j <= W;j++)
67 {
68 cin>>Map[i][j];
69 if(Map[i][j] == 'Y')
70 x1 = i,y1 = j;
71 if(Map[i][j] == 'M')
72 x2 = i,y2 = j;
73 }
74 bfs(x1,y1,0);
75 bfs(x2,y2,1);
76
77
78 /* for(int i =1;i <= L;i++)
79 {
80 for(int j = 1;j <= W;j++)
81 cout<<visit[0][i][j]<<" ";
82 cout<<endl;
83 }
84 cout<<endl;
85 for(int i =1;i <= L;i++)
86 {
87 for(int j = 1;j <= W;j++)
88 cout<<visit[1][i][j]<<" ";
89 cout<<endl;
90 }
91 */
92 long long min1 = L*W*4;
93 for(int i =1;i <= L;i++)
94 for(int j = 1;j <= W;j++)
95 {
96 if(Map[i][j] == '@'&&visit[0][i][j] != -1&&visit[1][i][j] != -1)
97 {
98 int sum = visit[0][i][j] + visit[1][i][j];
99 min1 = sum < min1?sum:min1;
100
101 }
102 }
103 cout<<min1*11<<endl;
104
105 }
106
107 return 0;
108 }