poj 3026 Borg Maze

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

题意：在一个y行 x列的迷宫中，有可行走的通路空格’ ‘，不可行走的墙’#’，还有两种英文字母A和S，现在从S出发，要求用最短的路径L连接所有字母，输出这条路径L的总长度；

思路：用bfs找出每两个点之间的距离，然后用prim求出最小生成树；思路是参考网上的；

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <queue>
using namespace std;
#define inf 2501
char map[51][51] = {'0'};
int node[51][51] = {0};
int col = 0;
int row = 0;
int num = 0;
int edge[102][102] = {0};
int dist[102][102] = {0};
struct node
{
    int row;
    int col;
}mov[4]={{0,-1},{0,1},{-1,0},{1,0}};
void bfs(int i,int j)
{
    bool visit[51][51] = {0};
    queue<int>x;
    queue<int>y;
    //memset(visit,0,sizeof(visit));
     x.push(i);
    y.push(j);
    int tem = node[i][j];
   //printf("%d hhhh\n",tem);
   memset(dist,0,sizeof(dist));
   visit[i][j] = 1;
    while(!x.empty())
    {

        int x1 = x.front();
        int y1 = y.front();

        x.pop();
        y.pop();
        if(node[x1][y1])
        edge[tem][node[x1][y1]] = dist[x1][y1];
        //printf("%d  \n",edge[node[i][j]][node[x1][y1]]);
        for(int k = 0;k < 4; ++k)
        {
            int mx = x1 + mov[k].row;
            int my = y1 + mov[k].col;
            if(mx >= 1&&mx <= row&&my >= 1&&my <= col&&!visit[mx][my]&&map[mx][my] != '#')
            {
                x.push(mx);
                y.push(my);
                visit[mx][my] = 1;
                dist[mx][my] = dist[x1][y1] + 1;
                //printf("%d \n",dist[mx][my]);
            }

        }
    }
     //printf("JJJJJdd\n");
}
void prim()
{
   int di[102] = {0};
   bool vis[102] = {0};
  // printf("JFDk");
   for(int i = 1;i <= num; ++i)
   {
       di[i] = edge[1][i];
       vis[i] = 0;
   }
   int ans = 0;
   vis[1] = 1;
   for(int i =1;i < num; ++i)
   {
       int min = inf;
       int p = 0;
       for(int j = 2;j <= num; ++j)
       if(!vis[j] && di[j] < min)
       {
           min = di[j];
           p = j;
       }
       vis[p] = 1;
       ans = ans + min;
       //printf("%d %d \n",min,p);
       for(int j = 1;j <= num; ++j)
       if(!vis[j] && di[j] > edge[p][j])
       di[j] = edge[p][j];
   }
   printf("%d\n",ans);
}
int main()
{
    int test = 0;
    cin>>test;
    while(test--)
    {
        memset(node,0,sizeof(node));
        num = 0;
        cin>>col>>row;
        char temp[51] = {0};
        gets(temp);
        for(int i = 1;i <= row; ++i)
        {
            gets(map[i]);
            for(int j = 1;j <= col;++j)
            if(map[i][j] == 'A' || map[i][j] == 'S')
            node[i][j] = ++num;
        }
        for(int i = 1;i <= row; ++i)
        for(int j = 1;j <= col; ++j)
        if(node[i][j])
        bfs(i,j);
        //for(int i = 1;i <= num;++i)
        //for(int j = 1;j <= num; ++j)
        //printf("%d %d %d\n",edge[i][j],i,j);
        prim();
    }
    return 0;
}