POJ 3206 Borg Maze (BFS+Prim)

Borg Maze
 

大意:给你一个m*n的迷宫,可以上下左右的走,只能走空格或字母,求出将所有字母连通起来的最小耗费。

 

思路:先用BFS求出S到所有A的距离,再用Prim求最小生成树,求出最小耗费。这个题坑的不在题,是数据太坑了,在空格处理上没弄好,贡献了好几个WA和CE,看Discuss才知道很坑,最后用G++过了的代码,C++还RE,实在不知道说什么好了  =。=

 

POJ 3206 Borg Maze (BFS+Prim)
  1 #include <stdio.h>
  2 #include <iostream>
  3 #include <string.h>
  4 #define INF 0xfffffff
  5 using namespace std;
  6 
  7 char str[55][55];
  8 int Map[55][55];
  9 int dis[55];
 10 int dist[105][105];
 11 int edge[105][105];
 12 int num, n, m, p    ;
 13 
 14 int Move[4][2] = {{0, -1}, {0, 1}, {-1, 0}, {1, 0}};
 15 
 16 int min(int a, int b)
 17 {
 18     return a > b ? b : a;
 19 }
 20 
 21 void BFS(int i, int j)
 22 {
 23     int head = 0, tail = 0;
 24     int q_x[2550], q_y[2550];
 25     bool vis[55][55];
 26     memset(vis, false, sizeof(vis));
 27     memset(dist, 0, sizeof(dist));
 28     vis[i][j] = true;
 29     q_x[tail] = i;
 30     q_y[tail++] = j;
 31     while(head < tail)
 32     {
 33         int x = q_x[head];
 34         int y = q_y[head++];
 35         if(Map[x][y])
 36         {
 37             edge[Map[i][j]][Map[x][y]] = dist[x][y];
 38         }
 39         for(int t = 0; t < 4; t++)
 40         {
 41             int dx = x+Move[t][0];
 42             int dy = y+Move[t][1];
 43             if(dx >= 1 && dx <= m && dy >= 1 && dy <= n)
 44             {
 45                 if(!vis[dx][dy] && str[dx][dy] != #)
 46                 {
 47                     q_x[tail] = dx;
 48                     q_y[tail++] = dy;
 49                     vis[dx][dy] = true;
 50                     dist[dx][dy] = dist[x][y]+1;
 51                 }
 52             }
 53         }
 54     }
 55 }
 56 
 57 
 58 int Prim()
 59 {
 60     int Ans;
 61     int Min_ele, Min_node;
 62     for(int i = 1; i <= num; i++)
 63     {
 64         dis[i] = INF;
 65     }
 66     Ans = 0;
 67     int r = 1;
 68     for(int i = 1; i <= num-1; i++)
 69     {
 70         Min_ele = INF;
 71         dis[r] = -1;
 72         for(int j = 1; j <= num; j++)
 73         {
 74             if(dis[j] >= 0)
 75             {
 76                 dis[j] = min(dis[j], edge[r][j]);
 77                 if(dis[j] < Min_ele)
 78                 {
 79                     Min_ele = dis[j];
 80                     Min_node = j;
 81                 }
 82             }
 83         }
 84         r = Min_node;
 85         Ans += Min_ele;
 86     }
 87     return Ans;
 88 }
 89 
 90 void Solve()
 91 {
 92     int i, j;
 93     cin >> p;
 94     while(p--)
 95     {
 96         memset(Map, 0, sizeof(Map));
 97         num = 0;
 98         cin >> n >> m;
 99         char s[100];
100         gets(s);///这里太坑了
101         for(int i = 1; i <= m; i++)
102         {
103             gets(str[i]);
104             for(int j = 0; j < n; j++)
105             {
106                 if(str[i][j] == A || str[i][j] == S)
107                 {
108                     Map[i][j] = ++num;
109                 }
110             }
111         }
112         for(int i = 1; i <= m; i++)
113         {
114             for(int j = 1; j <= n; j++)
115             {
116                 if(Map[i][j])
117                 {
118                     BFS(i, j);
119                 }
120             }
121         }
122         printf("%d\n", Prim());
123     }
124 }
125 
126 int main()
127 {
128     Solve();
129 
130     return 0;
131 }
Borg Maze

POJ 3206 Borg Maze (BFS+Prim)

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