POJ-3026 Borg Maze（BFS+最小生成树）

 

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

 

题意：

求有字母的各个点都彼此连接起来的最短步数。

思路：

BFS+最小生成树，刚开始就是不知道该怎样下手写

既然每一个 S 或 A 处才可以分裂，并且只能让其中一个行走，求最小的路径长度，便是一棵包含所有 S 或 A 点的树。

那么，首先通过BFS枚举每一个点到其他点的最短路径长度，若距离小于初始值，则更新该距离，然后建立这样一个完全图，然后在这一个完全图中求最小生成树即可。

 

注意，该题在输入x,y之后可能会出现很多空格哦！用一般的%*c和getchar()都会失效，所以我们用gets(str)，故WA了好多次。

 

根据题意的“分离”规则，重复走过的路不再计算

因此当使用prim算法求L的长度时，根据算法的特征恰好不用考虑这个问题（源点合并很好地解决了这个问题），L就是最少生成树的总权值W

由于使用prim算法求在最小生成树，因此无论哪个点做起点都是一样的，（通常选取第一个点），因此起点不是S也没有关系。

 

 

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <string>
#include <math.h>
#include <algorithm>
#include <vector>
#include <stack>
#include <queue>
#include <set>
#include <map>
#include <sstream>
const int INF=0x3f3f3f3f;
typedef long long LL;
const int mod=1e9+7;
//const double PI=acos(-1);
#define Bug cout<<"---------------------"<<endl
const int maxn=1e5+10;
using namespace std;

int NT[4][2]={{1,0},{0,1},{-1,0},{0,-1}};
char MP[110][110];//存放迷宫 
int G[110][110];//存放图
int number[110][110];//存放迷宫中S或A点所对应的编号 
int lowval[110];//辅助贪心数组 
int VIS[110][110];//BFS时判断迷宫的某个位置是否走过 
int vis[110];//Prim时判断顶点是否加入最小生成树 
int n,m;

struct node
{
    int x;
    int y;
    int step;
};

void BFS(int u,int x,int y)
{
    memset(VIS,0,sizeof(VIS));
    node begins,next;
    begins.x=x;
    begins.y=y;
    begins.step=0;
    queue<node> qe;
    qe.push(begins);
    VIS[x][y]=1;
    while(qe.size())
    {
        node now=qe.front();
        qe.pop();
        if(MP[now.x][now.y]=='A'||MP[now.x][now.y]=='S')
            G[u][number[now.x][now.y]]=now.step;
        for(int i=0;i<4;i++)
        {
            next.x=now.x+NT[i][0];
            next.y=now.y+NT[i][1];
            if(next.x>=0&&next.x<m&&next.y>=0&&next.y<n&&MP[next.x][next.y]!='#'&&!VIS[next.x][next.y])
            {
                VIS[next.x][next.y]=1;
                next.step=now.step+1;
                qe.push(next);
            }
        }
    }
}

int Prim(int N,int st)
{
    memset(vis,0,sizeof(vis));
    int sum=0;
    for(int i=0;i<N;i++)
        lowval[i]=G[st][i];
    vis[st]=1;
    for(int i=0;i<N-1;i++)
    {
        int t=-1;
        int MIN=INF;
        for(int j=0;j<N;j++)
        {
            if(!vis[j]&&lowval[j]<MIN)
            {
                MIN=lowval[j];
                t=j;
            }
        }
        sum+=MIN;
        vis[t]=1;
        for(int j=0;j<N;j++)
        {
            if(!vis[j]&&lowval[j]>G[t][j])
                lowval[j]=G[t][j];
        }
    }
    return sum;
} 


int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d %d ",&n,&m);//记得最后加上一个空格或者\n过滤掉输入中的空格 
        int k=0;
        for(int i=0;i<m;i++)
        {
            gets(MP[i]);
            for(int j=0;j<n;j++)
            {
                if(MP[i][j]=='A'||MP[i][j]=='S')
                    number[i][j]=k++;    
            }
        }
        for(int i=0;i<m;i++)
        {
            for(int j=0;j<n;j++)
            {
                if(i==j)
                    G[i][j]=0;
                else G[i][j]=INF;
            }
        }
        for(int i=0;i<m;i++)
        {
            for(int j=0;j<n;j++)
            {
                if(MP[i][j]=='A'||MP[i][j]=='S')
                    BFS(number[i][j],i,j);
            }
        }
        printf("%d\n",Prim(k,0));
    }
}