zoukankan      html  css  js  c++  java
  • POJ 3026 Borg Maze(bfs+最小生成树)

    Borg Maze
    Time Limit: 1000MS   Memory Limit: 65536K
    Total Submissions: 6634   Accepted: 2240

    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

     
     
     
     
     
    一开始一直看不懂题目意思。
     
    关键就是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.). 
     
     
    这样其实就是求最小生成树了
     
    先对有A或S的地方进行bfs。得出距离
     
    //============================================================================
    // Name        : POJ.cpp
    // Author      : 
    // Version     :
    // Copyright   : Your copyright notice
    // Description : Hello World in C++, Ansi-style
    //============================================================================
    
    #include <iostream>
    #include <stdio.h>
    #include <string.h>
    #include <algorithm>
    #include <queue>
    #include <map>
    using namespace std;
    
    char g[100][100];
    int n,m;
    int a[100][100];
    int move[][2]={{1,0},{-1,0},{0,1},{0,-1}};
    
    int cost[110][110];
    int t[100][100];
    void bfs(int sx,int sy)
    {
        queue<pair<int,int> >q;
        while(!q.empty())q.pop();
        memset(t,-1,sizeof(t));
        t[sx][sy]=0;
        q.push(make_pair(sx,sy));
        while(!q.empty())
        {
            pair<int,int> now=q.front();
            q.pop();
            if(a[now.first][now.second]!=-1)
                cost[a[sx][sy]][a[now.first][now.second]]=t[now.first][now.second];
            for(int i=0;i<4;i++)
            {
                int tx=now.first+move[i][0];
                int ty=now.second+move[i][1];
                if(g[tx][ty]=='#'||t[tx][ty]!=-1)continue;
                t[tx][ty]=t[now.first][now.second]+1;
                q.push(make_pair(tx,ty));
            }
        }
    }
    const int INF=0x3f3f3f3f;
    bool vis[110];
    int lowc[110];
    int Prim(int n)
    {
        int ans=0;
        memset(vis,false,sizeof(vis));
        vis[0]=true;
        for(int i=1;i<n;i++)lowc[i]=cost[0][i];
        for(int i=1;i<n;i++)
        {
            int minc=INF;
            int p=-1;
            for(int j=0;j<n;j++)
                if(!vis[j]&&minc>lowc[j])
                {
                    minc=lowc[j];
                    p=j;
                }
            if(minc==INF)return -1;
            ans+=minc;
            vis[p]=true;
            for(int j=0;j<n;j++)
                if(!vis[j]&&lowc[j]>cost[p][j])
                    lowc[j]=cost[p][j];
        }
        return ans;
    }
    
    
    int main()
    {
    //    freopen("in.txt","r",stdin);
    //    freopen("out.txt","w",stdout);
        int T;
        scanf("%d",&T);
        while(T--)
        {
            scanf("%d%d",&m,&n);
            gets(g[0]);
            memset(a,-1,sizeof(a));
            int tol=0;
            for(int i=0;i<n;i++)
            {
                gets(g[i]);
                for(int j=0;j<m;j++)
                    if(g[i][j]=='A'||g[i][j]=='S')
                        a[i][j]=tol++;
            }
            for(int i=0;i<n;i++)
                for(int j=0;j<n;j++)
                    if(a[i][j]!=-1)
                        bfs(i,j);
            printf("%d
    ",Prim(tol));
        }
        return 0;
    }
     
     
  • 相关阅读:
    Anaconda和Pycharm的安装和配置
    使用XAMPP集成开发环境安装Apache、PHP的配置说明
    新兴内存技术准备突围
    使嵌入式系统调试更容易:有用的硬件和软件提示
    保护嵌入式802.11 Wi-Fi设备时需要考虑的10件事
    关键任务应用程序依赖于故障保护存储器
    模拟内存计算如何解决边缘人工智能推理的功耗挑战
    如何为嵌入式应用选择适当的SSD
    Imec推出高性能芯片的低成本冷却解决方案
    交换机应用寻找10个完美的因素
  • 原文地址:https://www.cnblogs.com/kuangbin/p/3147031.html
Copyright © 2011-2022 走看看