POJ3026 Borg Maze 2017-04-21 16:02 50人阅读 评论(0) 收藏

Borg Maze

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 14165		Accepted: 4619

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

—————————————————————————————————————

题意：在一个迷宫里，起点是S，A代表外星人，我们需要找到最短的路径将S和所有的A连接起来，输出最短的这个路径总长

思路：先bfs建图，再最小生成树处理

注意：测试数据多了莫名其妙的空格，getchar（）会wa，看了discuss在知道，智商题浪费时间

#include <iostream>
#include<queue>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<set>
#include<cstring>
using namespace std;
#define LL long long

struct node
{
    int u,v,w;
} p[1000005];

struct point
{
    int x,y,t;
};

char mp[105][105];
int dir[4][2]= {{0,1},{0,-1},{1,0},{-1,0}};
int vis[105][105];

int n,m,cnt,tot,pre[10005];

bool cmp(node a,node b)
{
    return a.w<b.w;
}
void init()
{
    for(int i=0; i<10005; i++)
        pre[i]=i;
}

int fin(int x)
{
    return pre[x]==x?x:pre[x]=fin(pre[x]);
}

void kruskal()
{
    sort(p,p+cnt,cmp);
    init();
    int cost=0;
    int ans=0;
    for(int i=0; i<cnt; i++)
    {
        int a=fin(p[i].u);
        int b=fin(p[i].v);
        if(a!=b)
        {
            pre[a]=b;
            cost+=p[i].w;
            ans++;
        }
        if(ans==tot-1)
        {
            break;
        }
    }
    printf("%d\n",cost);
}

bool check(int x,int y)
{
    if(x<0||x>=n||y<0||y>=m||mp[x][y]=='#')
        return 0;
    return 1;
}

void build(int x,int y)
{
    memset(vis,0,sizeof vis);
    point f,d;
    f.x=x;
    f.y=y;
    f.t=0;
    vis[x][y]=1;
    queue<point>q;
    q.push(f);
    while(!q.empty())
    {
        f=q.front();
        q.pop();
        if((mp[f.x][f.y]=='A'||mp[f.x][f.y]=='S')&&(f.x!=x||f.y!=y))
        {
            p[cnt].u=100*x+y;
            p[cnt].v=100*f.x+f.y;
            p[cnt++].w=f.t;
        }
        for(int i=0; i<4; i++)
        {
            int xx=f.x+dir[i][0];
            int yy=f.y+dir[i][1];
            if(!vis[xx][yy]&&check(xx,yy))
            {
                d.x=xx;
                d.y=yy;
                d.t=f.t+1;
                vis[xx][yy]=1;
                q.push(d);
            }
        }
    }
}
int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d",&m,&n);
        char temp[51];
        gets(temp);
        for(int i=0; i<n; i++)
            gets(mp[i]);
        cnt=0,tot=0;
        for(int i=0; i<n; i++)
            for(int j=0; j<m; j++)
                if(mp[i][j]=='A'||mp[i][j]=='S')
                    build(i,j),tot++;
        kruskal();
    }
    return 0;
}