POJ 3026 Borg Maze

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本文介绍了一种用于估计迷宫搜索成本的算法，该算法帮助Borg集体找到捕获隐藏在迷宫中的外星人的最短路径。通过广度优先搜索计算外星人间的距离，并使用Prim最小生成树算法确定总搜索成本。

Borg Maze

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 12469		Accepted: 4083

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

Source

Svenskt Mästerskap i Programmering/Norgesmesterskapet 2001

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刚做的POJ，难度果然不一样（虽然是初级的233暑假集训的第一道ac题，这道题被列到了最小生成树，要是事先用最小生成树的话我估计根本想不到吧，很久没有碰代码了，最小生成树差点不会敲了都，而且这道题难点其实并不在最小生成树，关键是要找出A之间的距离，只要找出距离的话直接套用Prime就能做出来，而A之间的距离其实也很容易想到搜索，这种迷宫的两点最短距离直接可以用广搜，代码如下

Source Code

Problem: 3026		User: 15110572012
Memory: 256K		Time: 688MS
Language: C++		Result: Accepted

Source Code

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#define ll long long
#define inf 99999999
using namespace std;
int m[100+5][100+5];
int next[4][2]={{1,0},{0,1},{0,-1},{-1,0}};
bool vis[50+5][50+5];
struct A
{
    int x, y;
}A[100+5];
struct node
{
    int x, y;
    int step;
}boy[100000000+5];
int main()
{
    int n;
    scanf("%d", &n);
    while(n--)
    {
        int x, y, ans=0, ex, ey;
        scanf("%d %d\n", &x, &y);
        char s[50+5][50+5];
        for(int i=0;i<y;i++)
            gets(s[i]);
        for(int i=0;i<y;i++)
            for(int j=0;j<x;j++)
        {
            if(s[i][j]=='A')
            {
                A[ans].x=i;
                A[ans++].y=j;
            }
            else if(s[i][j]=='S')
            {
                ex=i;
                ey=j;
            }
        }
        A[ans].x=ex;
        A[ans++].y=ey;
        for(int i=0;i<ans;i++)
            for(int j=0;j<ans;j++)
        {
            if(i==j) m[i][j]=0;
            else
            {
                bool f=0;
                memset(vis,0,sizeof(vis));
                int head=0, tail=0;
                boy[tail].x=A[i].x;
                boy[tail].y=A[i].y;
                boy[tail++].step=0;
                vis[A[i].x][A[i].y]=1;
                while(head<tail)
                {
                    for(int i=0;i<4;i++)
                    {
                        int tx=boy[head].x;
                        int ty=boy[head].y;
                        tx+=next[i][0];
                        ty+=next[i][1];
                        if(tx>=0&&tx<x&&ty>=0&&ty<y&&!vis[tx][ty]&&s[tx][ty]!='#')
                        {
//                            cout<<tx<<' '<<ty<<' '<<s[tx][ty]<<endl;
                            vis[tx][ty]=1;
                            boy[tail].x=tx;
                            boy[tail].y=ty;
                            boy[tail++].step=boy[head].step+1;
                        }
                        if(boy[tail-1].x==A[j].x&&boy[tail-1].y==A[j].y)
                        {
                            f=1;
                            break;
                        }
                    }
                    if(f) break;
                    head++;
                }
                if(f)
                  m[i][j]=boy[tail-1].step;
                else m[i][j]=inf;
            }
        }
        int dis[1000+5];
        for(int i=0;i<ans;i++)
            dis[i]=m[0][i];
        bool vis[1000+5];
        memset(vis,0,sizeof(vis));
        vis[0]=1;
        int sum=0;
        for(int i=1;i<ans;i++)
        {
            int pos=-1, mi=inf;
            for(int j=0;j<ans;j++)
            if(dis[j]<mi&&!vis[j]) {pos=j;mi=dis[j];}
            if(pos==-1) break;
            sum+=mi;
            vis[pos]=1;
            for(int j=0;j<ans;j++)
                if(!vis[j]&&dis[j]>m[pos][j])
                   dis[j]=m[pos][j];
        }
        printf("%d\n", sum);
    }
    return 0;
}