poj 3026 Borg Maze

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

811

该题题义是某个人从S点出发，去寻找所有的A，他可以直接到达每个A，也可以通过分身来到达，具体视那种方法所走的路程短而定。换句话说就是可以从A点再走到A点来寻找下一个A，而不选择再从S出发。

首先将任意两点之间（A或者S）的距离求出来（通过BFS）然后再建立最小生成树即可。注意输入数据中x,y后面不只一个空格。

#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <cstdio>
#include <queue>

using namespace std;
#define N 100
#define M 100
const int inf=10000;
int map[N][N];
char str[N][N];
int row,col;          //行和列
int dx[4]={0,0,-1,1};
int dy[4]={1,-1,0,0};

struct node
{
  int x,y,t;
  friend bool operator<(node a,node b)
  {                            //距离短的优先级高
    return a.t>b.t;
  }
}f[M];

int judge(int x,int y)
{
  if(x>=0&&x<=row&&y>=0&&y<=col&&str[x][y]!='#')
    return 1;
  else
    return 0;
}

void bfs(int index,int n)
{
  int i,di,dj;
  int mark[M][M];        //标记该点是否在队列
  priority_queue<node>q;
  node cur,next;
  cur.x=f[index].x;
  cur.y=f[index].y;
  cur.t=0;
  q.push(cur);
  memset(mark,0,sizeof(mark));
  mark[cur.x][cur.y]=1;
  while(!q.empty())
  {
    cur=q.top();
    q.pop();
    for(i=0;i<n;i++)
    {
      if(cur.x==f[i].x&&cur.y==f[i].y)
      {
        map[index][i]=map[i][index]=cur.t;
        break;
      }
    }
    for(i=0;i<4;i++)
    {
      next.x=di=cur.x+dx[i];
      next.y=dj=cur.y+dy[i];
      next.t=cur.t+1;
      if(judge(di,dj)&&!mark[di][dj])
      {
        mark[di][dj]=1;
        q.push(next);
      }
    }
  }
}

int prim(int n)
{
  int i,index,min,sum=0;
  int dis[M],mark[M];
  for(i=0;i<n;i++)
    dis[i]=map[0][i];
  memset(mark,0,sizeof(mark));
  mark[0]=1;
  while(1)
  {
    index=-1;
    min=inf;
    for(i=0;i<n;i++)
      if(!mark[i]&&dis[i]<min)
      {
        min=dis[i];
        index=i;
      }
    if(index==-1)
      return sum;
    mark[index]=1;
    sum+=min;
    for(i=0;i<n;i++)
      if(!mark[i]&&dis[i]>map[index][i])
        dis[i]=map[index][i];
  }
  return 0;
}
int main()
{
  int T,i,j,n;
  char temp[200];
  scanf("%d",&T);
  while(T--)
  {
    scanf("%d%d",&col,&row);
    gets(temp);
    for(i=0;i<row;i++)
      gets(str[i]);
    for(i=0,n=0;i<row;i++)
      for(j=0;j<col;j++)
      {
        if(str[i][j]=='S'||str[i][j]=='A')
        {
          f[n].x=i;
          f[n++].y=j;
        }
      }
    for(i=0;i<n;i++)
      bfs(i,n);
    printf("%d\n",prim(n));
  }
  return 0;
}