poj 3026 Borg Maze (BFS + Prim)

#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[4][2] = {{1, 0}, {0, 1}, {-1, 0}, {0, -1}};

void Init()
{
    int i, j;
    memset(p, 0, sizeof(p));
    memset(vis, false, sizeof(vis));
    for(i = 1 ; i <= num ; i++)
    {
        for(j = 1 ; j <= num ; j++)
        {
            if(i == j)
                G[i][j] = 0;
            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 = 0;
    for(i = 1 ; i <= num ; i++)
        dist[i] = G[s][i];
    vis[s] = true;
    for(i = 1 ; i < num ; i++)
    {
        Min = INF;
        for(j = 1 ; j <= num ; j++)
        {
            if(!vis[j] && dist[j] < Min)
            {
                Min = dist[j];
                index = j;
            }
        }
        ans += Min;
        vis[index] = true;
        for(j = 1 ; 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 = 0;
    use[x][y] = true;
    Q.push(now);
    while(!Q.empty())
    {
        now = Q.front();
        Q.pop();
        if(p[now.x][now.y] > 0)//该点为字母
            G[p[x][y]][p[now.x][now.y]] = now.step;
        for(i = 0 ; i < 4 ; i++)
        {
            a = next.x = now.x + d[i][0];
            b = next.y = now.y + d[i][1];

            if(a >= 0 && a < m && b >= 0 && b < n && !use[a][b] && maps[a][b] != ‘#‘)
            {
                next.step = now.step + 1;
                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 = 0;
        for(i = 0 ; i < m ; i++)
        {
            for(j = 0 ; j < n ; j++)
            {
                scanf("%c", &maps[i][j]);
            }
            getchar();
        }
        for(i = 0 ; i < m ; i++)
        {
            for(j = 0 ; 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 = 0 ; i < m ; i++)
        {
            for(j = 0 ; j < n ; j++)
                if(p[i][j] > 0)
                    BFS(i, j);
        }
        printf("%d\n", prim(1));
    }
    return 0;
}