poj 3026 Borg Maze (BFS + Prim)

http://poj.org/problem?id=3026

Borg Maze

Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

Submit Status Practice POJ 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
最小生成树问题，但较不同点的是没有给原图，所以要用BFS找各个字母之间的距离，即构成原图

#include<stdio.h>
#include<string.h>
#include<math.h>
#include<stdlib.h>
#include<ctype.h>
#include<queue>
#define INF 0x3f3f3f3f
#define max(a, b)(a > b ? a : b)
#define min(a, b)(a < b ? a : b)
#define N 110

using namespace std;

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

int G[N][N], dist[N], p[N][N];//p记录字母的坐标
bool vis[N], use[N][N];
char maps[N][N];
int m, n, num;
int d[][] = {{, }, {, }, {-, }, {, -}};

void Init()
{
    int i, j;
    memset(p, , sizeof(p));
    memset(vis, false, sizeof(vis));
    for(i =  ; i <= num ; i++)
    {
        for(j =  ; j <= num ; j++)
        {
            if(i == j)
                G[i][j] = ;
            else
                G[i][j] = G[j][i] = INF;
        }
    }
}

int prim(int s)//最小生成树，求最小花费
{
    int i, j, index, Min, ans = ;
    for(i =  ; i <= num ; i++)
        dist[i] = G[s][i];
    vis[s] = true;
    for(i =  ; i < num ; i++)
    {
        Min = INF;
        for(j =  ; j <= num ; j++)
        {
            if(!vis[j] && dist[j] < Min)
            {
                Min = dist[j];
                index = j;
            }
        }
        ans += Min;
        vis[index] = true;
        for(j =  ; j <= num ; j++)
        {
            if(!vis[j] && dist[j] > G[index][j])
                dist[j] = G[index][j];
        }
    }
    return ans;
}

void BFS(int x, int y)//广搜构建原图
{
    queue<node>Q;
    int i, a, b;
    node now, next;
    memset(use, false, sizeof(use));
    now.x = x;
    now.y = y;
    now.step = ;
    use[x][y] = true;
    Q.push(now);
    while(!Q.empty())
    {
        now = Q.front();
        Q.pop();
        if(p[now.x][now.y] > )//该点为字母
            G[p[x][y]][p[now.x][now.y]] = now.step;
        for(i =  ; i <  ; i++)
        {
            a = next.x = now.x + d[i][];
            b = next.y = now.y + d[i][];

            if(a >=  && a < m && b >=  && b < n && !use[a][b] && maps[a][b] != '#')
            {
                next.step = now.step + ;
                use[a][b] = true;
                Q.push(next);

            }
        }
    }
}

int main()
{
    int i, j, t;
    scanf("%d", &t);
    while(t--)
    {
        scanf("%d%d ", &n, &m);
        Init();
        num = ;
        for(i =  ; i < m ; i++)
        {
            for(j =  ; j < n ; j++)
            {
                scanf("%c", &maps[i][j]);
            }
            getchar();
        }
        for(i =  ; i < m ; i++)
        {
            for(j =  ; j < n ; j++)
            {
                if(maps[i][j] == 'A' || maps[i][j] == 'S')
                    p[i][j] = ++num;//统计字母的个数，即要进入树的点的个数
            }
        }
        for(i =  ; i < m ; i++)
        {
            for(j =  ; j < n ; j++)
                if(p[i][j] > )
                    BFS(i, j);
        }
        printf("%d\n", prim());
    }
    return ;
}