hdoj 2612 find a way (两次bfs)

Find a way

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 6221    Accepted Submission(s): 2070

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
Y..@.
.#…
.#…
@..M.
#…#
 



Sample Output
66
88
66
 



Author
yifenfei
 



Source
 
 
一开始写成了让对没个肯德基店做一次bfs,会TLE
之后发现然对两个人分别做一次bfs就行
只是遇到肯德基店的 时候,因为不能确定这个是否是做优的,所以不return,继续搜下去.

作者: CrazyKK

ex-ACMer@hust,stackoverflow-engineer@sensetime

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