HDU 2612.Find A

Find A-to Time limit:1000MS Memory Limit:32768KB 64bit IO Format:%i64d &%i64 U Submit Status Practice HDU 2612

Description

Pass a year learning in Hangzhou, Yifenfei arrival hometown Ningbo at finally. Leave Ningbo One year, Yifenfei has many people to meet. Especially a good friend Merceki.
Yifenfei's home is on the countryside, but Merceki's home is in the center of the city. So Yifenfei made arrangements and Merceki to meet at a KFC. There is many KFC in Ningbo, they want to choose one and the total time to it is most smallest.
Now give your a Ningbo map, Both Yifenfei and Merceki can move up, down, left, right to the adjacent road by cost one minute S.

Input

The input contains multiple test cases.
Each test case include, first, 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, both Yifenfei and Merceki to arrival one of the KFC. Sure there is always has a KFC that can let them meet.

Sample Input

4 4y.#@.....#. @.. M4 4y.#@.....#. @#. M5 5[email protected].#....#...@. m.#...#

Sample Output

668866

There are two starting points of the BFS, each kfs is an exit, calculate the distance of each exit two inlet, calculate its minimum value to pay special attention to: KFC may not be able to reach

1 /*2 By:ohyee3 Github:ohyee4 Email:[email protected]5 Blog:http://www.cnblogs.com/ohyee/6 7 かしこいかわいい? 8 エリーチカ! 9 to write out the хорошо code OH ~Ten */ One  A#include <cstdio> -#include <algorithm> -#include <cstring> the#include <cmath> -#include <string> -#include <iostream> -#include <vector> +#include <list> -#include <queue> +#include <stack> A#include <map> at using namespacestd; -  - //DEBUG MODE - #defineDebug 0 -  - //Loops in #defineREP (n) for (int o=0;o<n;o++) -  to Const intMAXN =205; + intn,m; - CharMAP[MAXN][MAXN]; the Const intDelta[] = {1,-1,0,0}; *  $ structPoint {Panax Notoginseng     intx, y; - Point () { thex = y =-1; +     } APointintAintb) { thex =A; +y =b; -     } $ }; $ intnum; -  -  the intBFS (Point S,int(&dis) [MAXN] [MAXN]) { -memset (dis,-1,sizeof(DIS));WuyiQueue<point>Q; the  - Q.push (s); WuDIS[S.X][S.Y] =0; -  About      while(!Q.empty ()) { $         intx =Q.front (). x; -         inty =Q.front (). Y; - Q.pop (); -  AREP (4) { +             intxx = x +Delta[o]; the             intyy = y + delta[3-O]; -  $             if(XX <0|| XX >= N | | yy <0|| YY >=m) the                 Continue; the             if(Map[xx][yy] = ='#') the                 Continue; the             if(Dis[xx][yy] = =-1) { -DIS[XX][YY] = Dis[x][y] +1; in Q.push (Point (Xx,yy)); the             } the         } About     } the     return-1; the } the  + BOOLDo () { -     if(SCANF ("%d%d", &n,&m) = =EOF) the         return false;Bayi Point s1,s2; thenum =0; thePoint v[maxn*MAXN]; -      for(inti =0; I < n;i++) -          for(intj =0; J < m;j++) { thescanf"\n%c",&map[i][j]); the             if(Map[i][j] = ='Y') theS1 =Point (i,j); the             if(Map[i][j] = ='M') -S2 =Point (i,j); the             if(Map[i][j] = ='@')  thev[num++] =Point (i,j); the         }94  the     intDIS1[MAXN][MAXN]; the     intDIS2[MAXN][MAXN]; the 98 BFS (S1,DIS1); About BFS (S2,DIS2); - 101     intMin =100000;102      for(inti =0; I < num;i++)103         if(dis1[v[i].x][v[i].y]!=-1&& dis2[v[i].x][v[i].y]!=-1)104min = min (min,dis1[v[i].x][v[i].y]+dis2[v[i].x][v[i].y]); the 106printf"%d\n", Min * One);107 108     return true;109 } the 111 intMain () { the      while(Do ());113     return 0; the}

HDU 2612.Find A