POJ-3026 Borg Maze (BFS + minimum spanning tree)

 

http://poj.org/problem;jsessionid=A0F3392F460475E0058F93E009D6BFC3?id=3026

Description

The Borg is an immensely powerful race of enhanced humanoids from the delta quadrant of the galaxy. The Borg collective is the term used to describe the group consciousness of the Borg civilization. Each Borg individual is linked to the collective by a sophisticated subspace network that insures each member is given constant supervision and guidance. 

Your task is to help the Borg (yes, really) by developing a program which helps the Borg to estimate the minimal cost of scanning a maze for the assimilation of aliens hiding in the maze, by moving in north, west, east, and south steps. The tricky thing is that the beginning of the search is conducted by a large group of over 100 individuals. Whenever an alien is assimilated, or at the beginning of the search, the group may split in two or more groups (but their consciousness is still collective.). The cost of searching a maze is definied as the total distance covered by all the groups involved in the search together. That is, if the original group walks five steps, then splits into two groups each walking three steps, the total distance is 11=5+3+3.

Input

On the first line of input there is one integer, N <= 50, giving the number of test cases in the input. Each test case starts with a line containg two integers x, y such that 1 <= x,y <= 50. After this, y lines follow, each which x characters. For each character, a space `` '' stands for an open space, a hash mark ``#'' stands for an obstructing wall, the capital letter ``A'' stand for an alien, and the capital letter ``S'' stands for the start of the search. The perimeter of the maze is always closed, i.e., there is no way to get out from the coordinate of the ``S''. At most 100 aliens are present in the maze, and everyone is reachable.

Output

For every test case, output one line containing the minimal cost of a succesful search of the maze leaving no aliens alive.

Sample Input

2
6 5
##### 
#A#A##
# # A#
#S  ##
##### 
7 7
#####  
#AAA###
#    A#
# S ###
#     #
#AAA###
#####  

Sample Output

8
11

 

Meaning of the questions:

Seeking the number of steps each point shortest letters are connected to each other together.

Ideas:

BFS + minimum spanning tree, beginning just do not know how handwriting

Since each of the S or A can split, so that one can walk and, seeking the path of least length, that is a tree containing all S or A point.

So, BFS enumerated first path length through each point to the other point, if the distance is less than the initial value, the distance is updated, and the establishment of such a complete graph, and this minimum spanning tree to complete a drawing.

 

Note that this problem may occur after a lot of space Oh input x, y! Will fail with the general% * c and getchar (), so we use gets (str), so WA several times.

 

According to "separation" rule meaning of the title, walked the streets no longer repeat the calculation

Therefore, when using prim algorithm for the length L, according to the characteristics of the algorithm just do not consider the question (source combined solves this problem), L is the total minimum spanning tree weight value W

Due to the use of prim algorithm for minimum spanning tree, so no matter what point do the starting point is the same, (usually select the first point), and therefore is not a starting point S does not matter.

 

 

  1 #include <stdio.h>
  2 #include <string.h>
  3 #include <iostream>
  4 #include <string>
  5 #include <math.h>
  6 #include <algorithm>
  7 #include <vector>
  8 #include <stack>
  9 #include <queue>
 10 #include <set>
 11 #include <map>
 12 #include <sstream>
 13 const int INF=0x3f3f3f3f;
 14 typedef long long LL;
 15 const int mod=1e9+7;
 16 //const double PI=acos(-1);
 17 #define Bug cout<<"---------------------"<<endl
 18 const int maxn=1e5+10;
 19 using namespace std;
 20 
 21 int NT[4][2]={{1,0},{0,1},{-1,0},{0,-1}};
 22 char MP[110 ] [ 110 ]; // store labyrinth 
23 is  int G [ 110 ] [ 110 ]; // store FIG. 
24  int Number [ 110 ] [ 110 ]; // store S or A maze number corresponding point 
25  int lowval [ 110 ]; // auxiliary greedy array 
26 is  int VIS [ 110 ] [ 110 ]; // Analyzing maze when a location is passed BFS 
27  int VIS [ 110 ]; // when Prim minimum spanning tree is determined whether to join the vertex 
28  int n-, m;
 29 
 30 struct node
 31 {
 32     int x;
 33     int y;
 34     int step;
 35 };
 36 
 37 void BFS(int u,int x,int y)
 38 {
 39     memset(VIS,0,sizeof(VIS));
 40     node begins,next;
 41     begins.x=x;
 42     begins.y=y;
 43     begins.step=0;
 44     queue<node> qe;
 45     qe.push(begins);
 46     VIS[x][y]=1;
 47     while(qe.size())
 48     {
 49         node now=qe.front();
 50         qe.pop();
 51         if(MP[now.x][now.y]=='A'||MP[now.x][now.y]=='S')
 52             G[u][number[now.x][now.y]]=now.step;
 53         for(int i=0;i<4;i++)
 54         {
 55             next.x=now.x+NT[i][0];
 56             next.y=now.y+NT[i][1];
 57             if(next.x>=0&&next.x<m&&next.y>=0&&next.y<n&&MP[next.x][next.y]!='#'&&!VIS[next.x][next.y])
 58             {
 59                 VIS[next.x][next.y]=1;
 60                 next.step=now.step+1;
 61                 qe.push(next);
 62             }
 63         }
 64     }
 65 }
 66 
 67 int Prim(int N,int st)
 68 {
 69     memset(vis,0,sizeof(vis));
 70     int sum=0;
 71     for(int i=0;i<N;i++)
 72         lowval[i]=G[st][i];
 73     vis[st]=1;
 74     for(int i=0;i<N-1;i++)
 75     {
 76         int t=-1;
 77         int MIN=INF;
 78         for(int j=0;j<N;j++)
 79         {
 80             if(!vis[j]&&lowval[j]<MIN)
 81             {
 82                 MIN=lowval[j];
 83                 t=j;
 84             }
 85         }
 86         sum+=MIN;
 87         vis[t]=1;
 88         for(int j=0;j<N;j++)
 89         {
 90             if(!vis[j]&&lowval[j]>G[t][j])
 91                 lowval[j]=G[t][j];
 92         }
 93     }
 94     return sum;
 95 } 
 96 
 97 
 98 int main()
 99 {
100     int T;
101     scanf("%d",&T);
102     while(T--)
103     {
104         Scanf ( " % D% D " , & n-, & m); // remember the last with a space or \ n filtered out in the input space 
105          int K = 0 ;
 106          for ( int I = 0 ; I <m; I ++ )
 107          {
 108              the gets (the MP [I]);
 109              for ( int J = 0 ; J <n-; J ++ )
 110              {
 111                  IF (the MP [I] [J] == ' A ' || the MP [I] [ J] == ' S ' )
 112                     number[i][j]=k++;    
113             }
114         }
115         for(int i=0;i<m;i++)
116         {
117             for(int j=0;j<n;j++)
118             {
119                 if(i==j)
120                     G[i][j]=0;
121                 else G[i][j]=INF;
122             }
123         }
124         for(int i=0;i<m;i++)
125         {
126             for(int j=0;j<n;j++)
127             {
128                 if(MP[i][j]=='A'||MP[i][j]=='S')
129                     BFS(number[i][j],i,j);
130             }
131         }
132         printf("%d\n",Prim(k,0));
133     }
134 }