hdu 2612

Find a way

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3451    Accepted Submission(s):
1128

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

 

能不能用交替进行的方式进行双向广搜？

不行，因为可能你走的可能不是通往最佳的@的。

另加一组数据

6 5
Y.@#.
.#.#.
.#.#.
.#.#.
.#.#.
##M..

77

 #include<iostream>
 #include<stdio.h>
 #include<cstring>
 #include<cstdlib>
 #include<queue>
 using namespace std;

 int n,m;
 int yx,yy,mx,my;
 int to[][]={{,},{,},{-,},{,-}};
 char a[][];
 int  time[][][];
 bool hash[][][];
 bool flag;
 struct node
 {
     int x,y;
     int time;
 };
 queue<node>Q[];

 int Min(int x,int y)
 {
     return x>y? y:x;
 }
 bool pd(node &t)
 {
     if(t.x>=&&t.x<=n && t.y>=&&t.y<=m && a[t.x][t.y]!='#')return false;
     return true;
 }
 int bfs(int x)
 {
     int i,hxl=;
     node t,cur;

     while(!Q[x].empty())
     {
         cur=Q[x].front();
         Q[x].pop();
         for(i=;i<;i++)
         {
             t=cur;
             t.x=t.x+to[i][];
             t.y=t.y+to[i][];
             t.time++;
             if(pd(t))continue;
             if(hash[x][t.x][t.y])continue;
             hash[x][t.x][t.y]=true;
             time[x][t.x][t.y]=t.time;
             if(x== && a[t.x][t.y]=='@')
             {
                 hxl=Min(hxl,time[x^][t.x][t.y]+time[x][t.x][t.y]);
             }
             Q[x].push(t);
         }
     }
     return hxl;
 }
 void dbfs()
 {
     int ans=;
     node t;
     t.x=yx;
     t.y=yy;
     t.time=;
     Q[].push(t);
     hash[][yx][yy]=true;

     t.x=mx;
     t.y=my;
     t.time=;
     Q[].push(t);
     hash[][mx][my]=true;

     bfs();
     ans = bfs();
     printf("%d\n",ans*);
 }
 int main()
 {
     int i,j;
     while(scanf("%d%d",&n,&m)>)
     {
         for(i=;i<=n;i++)
             scanf("%s",a[i]+);
         memset(hash,false,sizeof(hash));
         memset(time,,sizeof(time));
         for(i=;i<=n;i++)
             for(j=;j<=m;j++){
                 if(a[i][j]=='Y'){
                     yx=i;
                     yy=j;
                 }
                 else if(a[i][j]=='M'){
                     mx=i;
                     my=j;
                 }
             }
         while(!Q[].empty()){
             Q[].pop();
         }
         while(!Q[].empty()){
             Q[].pop();
         }
         flag=false;
         dbfs();
     }
     return ;
 }