hdu 2612 Find a way

Find a way

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 10474    Accepted Submission(s): 3421

[align=left]Problem Description[/align]
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.

 

[align=left]Input[/align]
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

 

[align=left]Output[/align]
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.
 

[align=left]Sample Input[/align]

4 4
Y.#@
....
.#..
@..M
4 4
Y.#@
....
.#..
@#.M
5 5
Y..@.
.#...
.#...
@..M.
#...#

 

[align=left]Sample Output[/align]

66
88
66

 

[align=left]Author[/align]
yifenfei

题意：两人约在KFC店见面，但市里有很多家KFC店，每人走路的方向有四种选择，每当从一条路走到另一条路需要花费11分钟，选择一家KFC店使得两人花费的总时间最小。(    BFS   )

#include<stdio.h>
#include<queue>
#include<string.h>
#define M 100000000
using namespace std;
int n,m,flag,dis[210][210][2];
int vis[210][210];
int dx[4]={1,-1,0,0},dy[4]={0,0,1,-1};  //代表四个方向
char s[210][210];
struct node
{
int x,y;
int step;
};
int go(int a,int b)
{
if(a>=0&&b>=0&&a<n&&b<m&&s[a][b]!='#')
return 1;
return 0;
}
void bfs(int x,int y)
{
queue<node>que;
node st,ed;
st.x=x;
st.y=y;
st.step=0;
vis[x][y]=1;
que.push(st);
while(!que.empty())
{
st=que.front();
que.pop();
if(s[st.x][st.y]=='@')
{
dis[st.x][st.y][flag]=st.step;   //不能加 return  @_@
}
for(int i=0;i<4;i++)
{
ed.x=st.x+dx[i];
ed.y=st.y+dy[i];
if(go(ed.x,ed.y)&&!vis[ed.x][ed.y])
{
vis[ed.x][ed.y]=1;      //标记已经走过的路线
ed.step=st.step+1;
que.push(ed);
}
}
}
int main()
{
while(~scanf("%d%d",&n,&m))
{
getchar();
for(int i=0;i<n;i++)
{
for(int j=0;j<m;j++)
{
scanf("%c",&s[i][j]);
dis[i][j][0]=dis[i][j][1]=M;
}
getchar();
}
for(int i=0;i<n;i++)
{
for(int j=0;j<m;j++)
{
if(s[i][j]=='Y')
{
flag=0;
memset(vis,0,sizeof(vis));
bfs(i,j);
}
else if(s[i][j]=='M')
{
flag=1;
memset(vis,0,sizeof(vis));
bfs(i,j);
}
}
}
int max=M;
for(int i=0;i<n;i++)
{
for(int j=0;j<m;j++)
{
if(s[i][j]=='@'&&max>dis[i][j][0]+dis[i][j][1])    //求最短的时间
max=dis[i][j][0]+dis[i][j][1];
}
}
printf("%d\n",max*11);
}
}