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...
1 #include <iostream> 2 #include <queue> 3 using namespace std; 4 5 struct postion 6 { 7 int i, j; 8 postion operator+(postion p) 9 { 10 postion t; 11 t.i = this->i+ p.i; 12 t.j = this->j+ p.j; 13 return t; 14 } 15 }; 16 17 postion dir[] = {1,0, -1,0, 0,1, 0,-1}; 18 19 int step[305][305]; 20 //bool used[305][305]; 21 queue<postion>que; 22 char map[305][305]; 23 postion start, target; 24 int M, N; 25 26 int main() 27 { 28 int i, j; 29 while (cin>>M>>N && M+N) 30 { 31 for (i=0; i<M; i++) 32 for (j=0; j<N; j++) 33 { 34 cin>>map[i][j]; 35 step[i][j] = INT_MAX; 36 if (map[i][j] == 'Y') 37 start.i = i, start.j = j; 38 else if (map[i][j] == 'T') 39 target.i = i, target.j = j; 40 } 41 //memset(used, false, sizeof(used)); 42 //memset(step, 0, sizeof(step)); 43 //used[start.i][start.j] = true; 44 step[start.i][start.j] = 0; 45 while (!que.empty())que.pop(); 46 que.push(start); 47 postion tmp, t; 48 int sp; 49 while (!que.empty()) 50 { 51 tmp = que.front(); 52 que.pop(); 53 for (i=0; i<4; i++) 54 { 55 t = tmp + dir[i]; 56 if (t.i<0 || t.i>=M || t.j<0 || t.j>=N || map[t.i][t.j] == 'R' || map[t.i][t.j] == 'S') 57 continue; 58 if (map[t.i][t.j] == 'E') 59 sp = 1; 60 if (map[t.i][t.j] == 'B') 61 sp = 2; 62 if (step[t.i][t.j] > step[tmp.i][tmp.j] + sp) 63 { 64 step[t.i][t.j] = step[tmp.i][tmp.j] + sp; 65 que.push(t); 66 } 67 } 68 } 69 if (step[target.i][target.j] == INT_MAX) 70 step[target.i][target.j] = -1; 71 cout<<step[target.i][target.j]<<endl; 72 } 73 return 0; 74 }
