（POJ 3026） Borg Maze 最小生成树+bfs

题目链接：http://poj.org/problem?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  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 =++.
Input

On the first line of input there is one integer, N <= , giving the number of test cases in the input. Each test case starts with a line containg two integers x, y such that  <= x,y <= . 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  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

#####
#A#A##
# # A#
#S  ##
##### 

#####
#AAA###
#    A#
# S ###
#     #
#AAA###
#####
Sample Output

题意：从S点开始出发的机器人要把所有的A点同化，同化一个A后，A会变成S，继续同化A，问最短路径把A同化完

方法：先对每个S点与A点编号，再用bfs对每个点搜索，算出从这个点到别的所有的点的距离，最后就是求最小生成树

#include<stdio.h>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<queue>
#include<math.h>
#include <stack>
using namespace std;
#define ll long long
#define INF 0x3f3f3f3f
#define met(a,b) memset(a,b,sizeof(a))
#define mod 2147493647
#define N 100
int dir[][]= {{,},{,},{,-},{-,}};
char str[][];
int Map[][],vis[],dis[],vi[][];
int s[][],n,m;
struct node
{
    int x,y,s;
};
void dfs(int x,int y)
{
    met(vi,);
    queue<node>Q;
    node q,p;
    q.x=x;
    q.y=y;
    q.s=;
    Q.push(q);
    vi[x][y]=;
    int f=s[x][y];
    while(Q.size())
    {
        q=Q.front();Q.pop();

        if(str[q.x][q.y]>='A' && str[q.x][q.y]<='Z')
        {
            int e=s[q.x][q.y];
            Map[f][e]=q.s;
        }
        for(int i=; i<; i++)
        {
            p.x=q.x+dir[i][];
            p.y=q.y+dir[i][];
            p.s=q.s+;
            if(p.x< || p.x>=m || p.y< || p.y>=n)
                continue;
            if(s[p.x][p.y]>= && str[p.x][p.y]!='#'&&!vi[p.x][p.y])
            {
                Q.push(p);
                vi[p.x][p.y]=;
            }
        }
    }
}
int prim(int nn)
{
    int ans=;
    for(int i=; i<=nn; i++)
    {
        dis[i]=Map[][i];
        vis[i]=;
    }
    vis[]=;
    for(int i=; i<nn; i++)
    {
        int an=INF,k;
        for(int j=; j<=nn; j++)
        {
            if(!vis[j] && an>dis[j])
                an=dis[k=j];
        }
        ans+=an;
        vis[k]=;
        for(int j=; j<=nn; j++)
        {
            if(!vis[j])
                dis[j]=min(dis[j],Map[k][j]);
        }
    }
    return ans;
}
int main()
{
    int t;

    scanf("%d",&t);
    while(t--)
    {
        scanf("%d %d ",&n,&m);
        int k=;
        met(s,);
        for(int i=; i<m; i++)
        {
            gets(str[i]);
            for(int j=; j<n; j++)
            {
                if(str[i][j]>='A' && str[i][j]<='Z')
                    s[i][j]=k++;
            }
        }
        for(int i=; i<m; i++)
        {
            for(int j=; j<=n; j++)
                if(str[i][j]>='A' && str[i][j]<='Z')
                    dfs(i,j);
        }
        printf("%d\n",prim(k-));
    }
    return ;
}