Pku oj 3026 Borg Maze(BFS+MST)

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.

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.

For every test case, output one line containing the minimal cost of a succesful search of the maze leaving no aliens alive.

2
6 5
##### 
#A#A##
# # A#
#S  ##
##### 
7 7
#####  
#AAA###
#    A#
# S ###
#     #
#AAA###
#####  

8
11

 

题意是问把S和A全部连接起来最少需要多少步，拿第二个样例举例子

 

A—A—A

       |

       |— — —A

      S

       |

       |

A—A—A

如图，每一个横线代表一步，所以共需要11步，看图很容易就看出来这是一颗最小生成树，如何建图是关键。先将字母编号，然后遍历图中每一个字母，以每个字母为起点BFS其他字母，因为是BFS，所以得到的两个字母间发的距离一定是最小的，这就是两个点之间的权值，用此方法建图，然后Kru算法求最小生成树，代码有点长，细节很多，debug了好久，也算是锻炼了代码能力

还有最坑爹的一点！！！n，m后面有很多空格，我不知道这些的空格意义是什么，不过估计意义就是出题人脑残，所以要记得用gets吸收多余的空格

附上一组大数据，过不了的同学可以用这组大数据测试一下：

1
50 50
##################################################
#AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA#
#                                               A#
#                                               A#
#                                               A#
#                                               A#
#                                               A#
#                                               A#
#                                               A#
#                                               A#
#                                               A#
#                                               A#
#                                               A#
#                                               A#
#                                               A#
#                                               A#
#                                               A#
#                                               A#
#                                               A#
#                                               A#
#                                               A#
#                                               A#
#                                               A#
#                                               A#
#                                               A#
#                                               A#
#                                               A#
#                                               A#
#                                               A#
#                                               A#
#                                               A#
#                                               A#
#                                               A#
#                                               A#
#                                               A#
#                                               A#
#                                               A#
#                                               A#
#                                               A#
#                                               A#
#                                               A#
#                                               A#
#                                               A#
#                                               A#
#                                               A#
#                                               A#
#                                               A#
#                                               A#
#S                                              A#
##################################################

答案应该输出141

最后上代码

#include<cstdio>
#include<iostream>
#include<cstring>
#include<queue>
#include<algorithm>
#include<cstring>
using namespace std;
const int maxn = 505;
int n,m;
int ans,cnt;
char space[1000];
int vis[maxn][maxn];
int fx[4] = {0,0,1,-1};
int fy[4] = {1,-1,0,0};
int set[maxn * maxn];
char a[maxn][maxn];
int b[maxn][maxn];

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

point now,temp,end;

struct Edge
{
    int a,b;
    int dis;
}edge[maxn * maxn];

void bfs(int x,int y,int from)
{
    memset(vis,0,sizeof(vis));
    queue <point> q;
    now.x = x;
    now.y = y;
    now.dis = 0;
    vis[x][y] = 1;
    q.push(now);
    while(!q.empty())
    {
        temp = q.front();
        q.pop();
        for(int p=0;p<4;p++)
        {
            end.x = temp.x + fx[p];
            end.y = temp.y + fy[p];
            end.dis = temp.dis + 1;
            if(end.x >=0 && end.x < n && end.y >= 0 && end.y < m && vis[end.x][end.y] == 0 && b[end.x][end.y] != -1)
            {
                vis[end.x][end.y] = 1;
                q.push(end);
                if(b[end.x][end.y] > 0)
                {
                    edge[ans].a = from;
                    edge[ans].b = b[end.x][end.y];
                    edge[ans++].dis = end.dis;
                }
            }
        }
    }
}

int cmp(Edge a,Edge b)
{
    return a.dis < b.dis;
}  //cmp

int find(int x)
{
    int k,j,r;
    r = x;
    while(r != set[r])
        r = set[r];
    k = x;
    while(k != r)
    {
        j = set[k];
        set[k] = r;
        k = j;
    }
    return r;
}  //并查集查找 + 路径压缩

void merge(int x,int y)
{
    set[y] = x;
} //并查集合并

void init()
{
    for(int i=1;i<=cnt;i++)
        set[i] = i;
} //并查集初始化；

int Kruskal()
{
    init();
    int ans1 = 0;
    sort(edge,edge + ans,cmp);
    for(int i=0;i<ans;i++)
    {
        int f1 = find(edge[i].a);
        int f2 = find(edge[i].b);
        if(f1 == f2)
            continue;
        else
        {
            ans1 += edge[i].dis;
            merge(f1,f2);
        }
    }
    return ans1;
}  //Kru算法

int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        memset(b,0,sizeof(b));
        scanf("%d%d",&m,&n);
        gets(space);
        for(int i=0;i<n;i++)
        {
            gets(a[i]);
        }
        cnt = 1;
        ans = 0;
        for(int i=0;i<n;i++)
        {
            for(int j=0;j<m;j++)
            {
                if(a[i][j] == '#')
                    b[i][j] = -1;
                if(a[i][j] == ' ')
                    b[i][j] = 0;
                if(a[i][j]=='S'||a[i][j]=='A')
                {
                    b[i][j] = cnt;
                    cnt++;
                }
            }
        }
        for(int i=0;i<n;i++)
        {
            for(int j=0;j<m;j++)
            {
                if(b[i][j] > 0)
                    bfs(i,j,b[i][j]);
            }
        }
        int flag = Kruskal();
        printf("%d\n",flag);
    }
    return 0;
}