POJ 3026 Borg Maze bfs+Kruskal

题目链接：http://poj.org/problem?id=3026

感觉英语比题目本身难，其实就是个最小生成树，不过要先bfs算出任意两点的权值。

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

 char maze[][];
 int par[];
 struct Point
 {
     int x, y;
 }point[];

 struct Edge
 {
     int u, v, w;
     bool operator<(const struct Edge &b)const
     {
         return w < b.w;
     }
 }edge[];

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

 int find_set(int x)
 {
     return x == par[x] ? x : par[x] = find_set(par[x]);
 }

 queue<struct node>q;
 bool vis[][];
 int bfs(int x, int y, int ex, int ey)
 {
     while(!q.empty())q.pop();
     memset(vis, , sizeof(vis));
     int dir[][] = {{, }, {, -}, {-, }, {, }};
     q.push((struct node){x, y, });
     vis[x][y] = ;
     while(!q.empty())
     {
         struct node u = q.front();
         q.pop();
         if(u.x == ex && u.y == ey)
             return u.step;
         for(int i = ; i < ; i++)
         {
             if(!vis[u.x+dir[i][]][u.y+dir[i][]] && maze[u.x+dir[i][]][u.y+dir[i][]] != '#')
             {
                 vis[u.x+dir[i][]][u.y+dir[i][]] = ;
                 q.push((struct node){u.x+dir[i][], u.y+dir[i][], u.step+});
             }
         }
     }
 }

 int main()
 {
     int t, n, m;
     char fuck_space[];
     scanf("%d%*c", &t);
     while(t--)
     {
         gets(fuck_space);
         sscanf(fuck_space, "%d %d", &m, &n);
         int point_rear = ;
         for(int i = ; i < n; i++)
         {
             gets(maze[i]);
             for(int j = ; j < m; j++)
                 if(maze[i][j] == 'S' || maze[i][j] == 'A')
                     point[point_rear++] = (struct Point){i, j};
         }
         int edge_rear = ;
         for(int i = ; i < point_rear; i++)
         {
             for(int j = i+; j < point_rear; j++)
             {
                 int w = bfs(point[i].x, point[i].y, point[j].x, point[j].y);
                 edge[edge_rear++] = (struct Edge){i, j, w};
             }
         }
         int ans = ;
         for(int i = ; i < point_rear; i++)
             par[i] = i;
         sort(edge, edge+edge_rear);
         for(int i = ; i < edge_rear; i++)
         {
             int x = find_set(edge[i].u);
             int y = find_set(edge[i].v);
             if(x != y)
             {
                 ans += edge[i].w;
                 par[x] = y;
             }
         }
         printf("%d\n", ans);
     }
     return ;
 }