poj3026 Borg Maze

Borg Maze

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 10424		Accepted: 3459

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

题目大意就是，将所有的S或者A点连起来的最小距离。

这题不得不吐槽，太坑。。。。 我TLE了一天最后是标记没有用BOOL,并且吃字符用getchar()RE，只能用gets(Map[0]);

思路： 将途中S与A用a[][]记录下位置，a[][]保存的值是从1--(S+A的数量 )=n，然后轮流以n个点作为顶点进行收索。构造与其他点间的距离，然后存在mt[][]中。

到其他点的距离实现的方法是用f.ans来记录。最后prim一下。

#include<iostream>
#include<cstring>
#include<cstdio>
#include<queue>
#define inf 0x3f3f3f3f
using namespace std;
struct node
{
    int x,y,ans;
};
char Map[2000][2000];
int n,m;
int a[2000][2000],point,mt[2000][2000],dis[2000];
int cx[]= {0,0,-1,1},cy[]= {1,-1,0,0},ans;
bool vis[2000][2000],bj[2000];//血的教训，以后什么标记都用bool!!!
void BFS(int xx,int yy)
{
    memset(vis,0,sizeof(vis));
    queue<node >q;
    node o,f;
    o.x=xx;
    o.y=yy;
    o.ans=0;
    q.push(o);
    vis[xx][yy]=1;
    while(!q.empty())
    {
        o=q.front();
        q.pop();
        if(a[o.x][o.y]!=0)
            mt[a[xx][yy]][a[o.x][o.y] ]=o.ans;
        for(int i=0; i<4; i++)
        {
            f=o;
            f.x+=cx[i];
            f.y+=cy[i];
            if(f.x>=0&&f.x<m&&f.y>=0&&f.y<n&&!vis[f.x][f.y]&&Map[f.x][f.y]!='#')
            {
                vis[f.x][f.y]=1;
                f.ans++;
                q.push(f);
            }
        }
    }
}
void prim()
{
ans=0;
    int i,j,pos,MAX;
    for(i=1; i<=point; i++)
        dis[i]=mt[1][i];
    dis[1]=0;
    memset(bj,0,sizeof(bj));
    bj[1]=1;
    for(i=1; i<point; i++)
    {
        MAX=inf;
        for(j=1; j<=point; j++)
        {
            if(MAX>dis[j]&&!bj[j])
            {
                MAX=dis[j];
                pos=j;
            }
        }
        ans+=MAX;
        bj[pos]=1;
        for(j=1; j<=point; j++)
        {
            if(dis[j]>mt[pos][j]&&!bj[j])
            {
                dis[j]=mt[pos][j];
            }
        }
    }
    printf("%d\n",ans);
}
int main()
{
    int i,j,k,cla;
    ios::sync_with_stdio(false);
    scanf("%d",&cla);
    while(cla--)
    {
        scanf("%d%d",&n,&m);
        point=0;
        memset(a,0,sizeof(a));
        gets(Map[0]);
        for(int i=0; i<m; i++)
        {
            gets(Map[i]);
            for(int j=0; j<n; j++)
            {
                if(Map[i][j] == 'S' || Map[i][j] == 'A')
                {
                    a[i][j] = ++point;
                }
            }
        }
        for(i=0; i<m; i++)
        {
            for(j=0; j<n; j++)
            {
                if(a[i][j]!=0)
                {
                    BFS(i,j);
                }
            }
        }
        prim();
    }
    return 0;
}