poj2312Battle City BFS

题意: M行N列矩阵, 'Y'表示开始位置, 'T'表示目标位置, 从开始位置到目标位置至少需要走多少步,其中, 'S', 'R'表示不能走, 'B' 花费为2, 'E'花费为1.

思路:纯 BFS.

很久没写BFS, 刚写居然不知道从何下手...

 #include <iostream>
 #include <queue>
 using namespace std;

 struct postion
 {
     int i, j;
     postion operator+(postion p)
     {
         postion t;
         t.i = this->i+ p.i;
         t.j = this->j+ p.j;
         return t;
     }
 };

 postion dir[] = {,, -,, ,, ,-};

 int step[][];
 //bool used[305][305];
 queue<postion>que;
 char map[][];
 postion start, target;
 int M, N;

 int main()
 {
     int  i, j;
     while (cin>>M>>N && M+N)
     {
         for (i=; i<M; i++)
             for (j=; j<N; j++)
             {
                 cin>>map[i][j];
                 step[i][j] = INT_MAX;
                 if (map[i][j] == 'Y')
                     start.i = i, start.j = j;
                 else if (map[i][j] == 'T')
                     target.i = i, target.j = j;
             }
             //memset(used, false, sizeof(used));
             //memset(step, 0, sizeof(step));
             //used[start.i][start.j] = true;
             step[start.i][start.j] = ;
             while (!que.empty())que.pop();
             que.push(start);
             postion tmp, t;
             int sp;
             while (!que.empty())
             {
                 tmp = que.front();
                 que.pop();
                 for (i=; i<; i++)
                 {
                     t = tmp + dir[i];
                     if (t.i< || t.i>=M || t.j< || t.j>=N || map[t.i][t.j] == 'R' || map[t.i][t.j] == 'S')
                         continue;
                     if (map[t.i][t.j] == 'E')
                         sp = ;
                     if (map[t.i][t.j] == 'B')
                         sp = ;
                     if (step[t.i][t.j] > step[tmp.i][tmp.j] + sp)
                     {
                         step[t.i][t.j] = step[tmp.i][tmp.j] + sp;
                         que.push(t);
                     }
                 }
             }
             if (step[target.i][target.j] == INT_MAX)
                 step[target.i][target.j] = -;
             cout<<step[target.i][target.j]<<endl;
     }
     return ;
 }