图论 --- BFS + MST

Borg
Maze

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

#include<queue>

#include<cstdio>

#define INF 1<<30

#include<cstring>

#include<algorithm>

using namespace std;

int n,m,ans,cnt;

struct Node

{

    int x,y,step;

};

Node node[110];

int row,line;

int dist[60];

char Map[60][60];

bool vis[60][60],vist[60]; //邻接矩阵存储

int mark[60][60],dis[110][110];     //给每个字母一个编号

int dir[4][2]={-1,0, 0,1, 1,0, 0,-1};
void bfs(int index)

{

    queue<Node> que;

    memset(vis, false, sizeof(vis));

    Node now, next;

    node[index].step = 0;

    que.push(node[index]);

    vis[node[index].x][node[index].y] = true;

    while(!que.empty())

    {

        now = que.front();

        que.pop();

        int x = now.x, y = now.y;

        for(int i = 0; i < 4; ++i)

        {

            int tx = x + dir[i][0], ty = y +dir[i][1];

            if(vis[tx][ty] == false && Map[tx][ty] != ‘#‘) //如果这一步可以走

            {

                next.x = tx;

                next.y = ty;

                vis[tx][ty] = true;

                next.step = now.step + 1;

                que.push(next);

                if(Map[next.x][next.y] == ‘A‘ || Map[next.x][next.y] == ‘S‘)

                    dis[ mark[ node[index].x ][ node[index].y ] ][ mark[next.x][next.y] ] = next.step;

            }
}

    }

}
int prim()

{

    for(int i = 0; i < cnt; ++i)

    {

        dist[i] = INF;

        vist[i] = false;

    }

    dist[0] = 0;

    while(1)

    {

        int min = INF, now = -1;

        for(int i = 0; i < cnt; ++i)

        {

            if(min > dist[i] && vist[i] == false)

            {

                min = dist[i];

                now = i;

            }

        }

        if(now == -1)

            return ans;

        ans += min;

        vist[now] = true;

        for(int i = 0; i < cnt; ++i)

            if(vist[i] == false && dist[i] > dis[now][i])

                dist[i] = dis[now][i];

    }

    return ans;

}
int main()

{

    int n_case;

    scanf("%d", &n_case);

    while(n_case--)

    {

        cnt = ans = 0;

        memset(mark, 0, sizeof(mark));

        scanf("%d%d", &row, &line);

        char ch;

        while(ch = getchar(), ch != ‘\n‘);

        for(int i = 0; i < line; ++i)

        {

            for(int j = 0; j < row; ++j)

            {

                Map[i][j] = getchar();

                if(Map[i][j] == ‘A‘ || Map[i][j] == ‘S‘)

                {

                    mark[i][j] = cnt;

                    node[cnt].x = i;

                    node[cnt++].y = j;

                }

            }

            while(ch = getchar(), ch != ‘\n‘);

        }

        for(int i = 0; i < cnt; ++i)

            bfs(i);

        prim();

        printf("%d\n", ans);

    }

    return 0;

}

图论 --- BFS + MST