POJ 3026 Borg Maze

#include<stdio.h>
#include<queue>
#include<string.h>
#include<algorithm>
using namespace std;

const int INF=0x3f3f3f3f;
const int N=110;

struct node
{
    int x, y, s;
};
int G[N][N], dist[N], vis[N], no[N][N], m, n, ans; ///no数组存放的是第(i，j)位置是A或S的第几个序号，ans代表S和A的总个数
int dir[4][2] = { {1,0}, {-1,0}, {0,1}, {0,-1} };
char Map[N][N];

void Init()
{
    int i, j;

    for (i = 0; i < N; i++)
    {
        dist[i] = INF;
        vis[i] = 0;
        for (j = 0; j < N; j++)
        {
            G[i][j] = INF;
            no[i][j] = 0;
        }
        G[i][i] = 0;
    }
    ans = 0;
}

void BFS(int x, int y, int z)
{
    int i, visit[N][N], nx, ny;
    node now, next;
    queue<node>Q;

    memset(visit, 0, sizeof(visit));

    now.x = x; now.y = y; now.s = 0;
    Q.push(now);
    visit[x][y] = 1;

    while (!Q.empty())
    {
        now = Q.front(); Q.pop();

        if (Map[now.x][now.y] == ‘S‘ || Map[now.x][now.y] == ‘A‘)
            G[z][no[now.x][now.y]] = G[no[now.x][now.y]][z] = now.s; ///遇到A和S，就保存要查询的节点到该点的最小距离（最小生成树是无向图）

        for (i = 0; i < 4; i++)
        {
            next.x = nx = now.x + dir[i][0];
            next.y = ny = now.y + dir[i][1];

            if (nx >= 0 && nx < m && ny >= 0 && ny < n && Map[nx][ny] != ‘#‘ && !visit[nx][ny])
            {
                visit[nx][ny] = 1;
                next.s = now.s+1;
                Q.push(next);
            }
        }
    }
}

int Prim()
{
    int i, j, Min, idex, num = 0;

    for (i = 1; i <= ans; i++)
        dist[i] = G[1][i];
    vis[1] = 1; ///！！！易掉

    for (i = 1; i < ans; i++)
    {
        Min = INF;

        for (j = 1; j <= ans; j++)
        {
            if (!vis[j] && dist[j] < Min)
            {
                Min = dist[j];
                idex = j;
            }
        }

        vis[idex] = 1;
        num += Min;

        for (j = 1; j <= ans; j++)
        {
            if (dist[j] > G[idex][j] && !vis[j])
                dist[j] = G[idex][j];
        }
    }

    return num;
}

int main ()
{
    int T, i, j, num;

    scanf("%d", &T);

    while (T--)
    {
        Init();

        scanf("%d%d ", &n, &m);
        for (i = 0; i < m; i++)
            gets(Map[i]);

        for (i = 0; i < m; i++)
            for (j = 0; j < n; j++)
                if (Map[i][j] == ‘S‘ || Map[i][j] == ‘A‘)
                    no[i][j] = ++ans; ///将该位置节点的序号记录下来

        for (i = 0; i < m; i++)
            for (j = 0; j < n; j++)
                if (Map[i][j] == ‘S‘ || Map[i][j] == ‘A‘)
                    BFS(i, j, no[i][j]); ///每次遇到A和S就进行BFS，查找其与其他所有的S和A的距离

        num = Prim();

        printf("%d\n", num);
    }

    return 0;
}