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POJ 3026 Borg Maze bfs+Kruskal

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。

  1 #include <stdio.h>
  2 #include <string.h>
  3 #include <queue>
  4 #include <algorithm>
  5 using namespace std;
  6 
  7 char maze[50][50];
  8 int par[50];
  9 struct Point
 10 {
 11     int x, y;
 12 }point[110];
 13 
 14 struct Edge
 15 {
 16     int u, v, w;
 17     bool operator<(const struct Edge &b)const
 18     {
 19         return w < b.w;
 20     }
 21 }edge[10000];
 22 
 23 struct node
 24 {
 25     int x, y, step;
 26 };
 27 
 28 int find_set(int x)
 29 {
 30     return x == par[x] ? x : par[x] = find_set(par[x]);
 31 }
 32 
 33 queue<struct node>q;
 34 bool vis[50][50];
 35 int bfs(int x, int y, int ex, int ey)
 36 {
 37     while(!q.empty())q.pop();
 38     memset(vis, 0, sizeof(vis));
 39     int dir[4][2] = {{0, 1}, {0, -1}, {-1, 0}, {1, 0}};
 40     q.push((struct node){x, y, 0});
 41     vis[x][y] = 1;
 42     while(!q.empty())
 43     {
 44         struct node u = q.front();
 45         q.pop();
 46         if(u.x == ex && u.y == ey)
 47             return u.step;
 48         for(int i = 0; i < 4; i++)
 49         {
 50             if(!vis[u.x+dir[i][0]][u.y+dir[i][1]] && maze[u.x+dir[i][0]][u.y+dir[i][1]] != '#')
 51             {
 52                 vis[u.x+dir[i][0]][u.y+dir[i][1]] = 1;
 53                 q.push((struct node){u.x+dir[i][0], u.y+dir[i][1], u.step+1});
 54             }
 55         }
 56     }
 57 }
 58 
 59 int main()
 60 {
 61     int t, n, m;
 62     char fuck_space[100];
 63     scanf("%d%*c", &t);
 64     while(t--)
 65     {
 66         gets(fuck_space);
 67         sscanf(fuck_space, "%d %d", &m, &n);
 68         int point_rear = 0;
 69         for(int i = 0; i < n; i++)
 70         {
 71             gets(maze[i]);
 72             for(int j = 0; j < m; j++)
 73                 if(maze[i][j] == 'S' || maze[i][j] == 'A')
 74                     point[point_rear++] = (struct Point){i, j};
 75         }
 76         int edge_rear = 0;
 77         for(int i = 0; i < point_rear; i++)
 78         {
 79             for(int j = i+1; j < point_rear; j++)
 80             {
 81                 int w = bfs(point[i].x, point[i].y, point[j].x, point[j].y);
 82                 edge[edge_rear++] = (struct Edge){i, j, w};
 83             }
 84         }
 85         int ans = 0;
 86         for(int i = 0; i < point_rear; i++)
 87             par[i] = i;
 88         sort(edge, edge+edge_rear);
 89         for(int i = 0; i < edge_rear; i++)
 90         {
 91             int x = find_set(edge[i].u);
 92             int y = find_set(edge[i].v);
 93             if(x != y)
 94             {
 95                 ans += edge[i].w;
 96                 par[x] = y;
 97             }
 98         }
 99         printf("%d
", ans);
100     }
101     return 0;
102 }

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原文地址：https://www.cnblogs.com/wolfred7464/p/3242392.html