标签:== href tmp 链接 ace include sizeof namespace span
题目链接
http://poj.org/problem?id=1475
题意
推箱子游戏。输入迷宫、箱子的位置、人的位置、目标位置,求人是否能把箱子推到目标位置,若能则输出推的最少的路径,如果有多条步数相同的推的最少的路径,则输出总步数(人走的步数+推箱子的步数)最少的那条路径;若不能把箱子推到目标位置,则输出Impossible.
思路
先求出箱子到目标位置的最短路径(bfs_box),在bfs1推箱子的过程中,根据推的方向和箱子的位置得到人的位置,再求得人到达这个位置的最短路(bfs_person)即可。
代码
1 #include <iostream> 2 #include <cstring> 3 #include <cstdio> 4 #include <string> 5 #include <queue> 6 using namespace std; 7 8 struct Node 9 { 10 int br, bc; //box_row,box_col 11 int pr, pc; //person_row,person_col 12 string ans; 13 14 Node() {} 15 Node(int br, int bc, int pr, int pc, string ans) :br(br), bc(bc), pr(pr), pc(pc), ans(ans) {} 16 }; 17 18 const int N = 20; 19 int m, n; 20 char maze[N][N]; 21 int dir[4][2] = { {-1, 0}, {1, 0}, {0, -1}, {0, 1} }; 22 char push[] = { ‘N‘, ‘S‘, ‘W‘, ‘E‘ }; 23 char walk[] = { ‘n‘, ‘s‘, ‘w‘, ‘e‘ }; 24 int br, bc; 25 int pr, pc; 26 int tr, tc; 27 28 bool ok(int r, int c) 29 { 30 if (r >= 0 && r < m && c >= 0 && c < n && maze[r][c] != ‘#‘) 31 return true; 32 return false; 33 } 34 35 string tmp; 36 bool bfs_person(int sr, int sc, int er, int ec, Node node) 37 { 38 tmp = ""; 39 int visit[N][N]; 40 memset(visit, 0, sizeof(visit)); 41 queue<Node> q; 42 q.push(Node(-1, -1, sr, sc, "")); 43 visit[sr][sc] = 1; 44 visit[node.br][node.bc] = 1; //注意 45 while (!q.empty()) 46 { 47 Node cur = q.front(); 48 q.pop(); 49 if (cur.pr == er && cur.pc == ec) 50 { 51 tmp = cur.ans; 52 return true; 53 } 54 for (int i = 0; i < 4; i++) 55 { 56 int nr = cur.pr + dir[i][0]; 57 int nc = cur.pc + dir[i][1]; 58 if (ok(nr, nc) && !visit[nr][nc]) 59 { 60 visit[nr][nc] = 1; 61 string ans = cur.ans + walk[i]; 62 q.push(Node(-1, -1, nr, nc, ans)); 63 } 64 } 65 } 66 return false; 67 } 68 69 string bfs_box() 70 { 71 int visit[N][N]; 72 memset(visit, 0, sizeof(visit)); 73 queue<Node> q; 74 q.push(Node(br, bc, pr, pc, "")); 75 visit[br][bc] = 1; 76 while (!q.empty()) 77 { 78 Node cur = q.front(); 79 q.pop(); 80 if (cur.br == tr && cur.bc == tc) 81 return cur.ans; 82 for (int i = 0; i < 4; i++) 83 { 84 int nr = cur.br + dir[i][0]; 85 int nc = cur.bc + dir[i][1]; 86 int pre_r = cur.br - dir[i][0]; 87 int pre_c = cur.bc - dir[i][1]; 88 if (ok(nr, nc) && ok(pre_r, pre_c) && !visit[nr][nc]) 89 { 90 if (bfs_person(cur.pr, cur.pc, pre_r, pre_c, cur)) 91 { 92 visit[nr][nc] = 1; 93 Node next; 94 next.br = nr; 95 next.bc = nc; 96 next.pr = cur.br; 97 next.pc = cur.bc; 98 next.ans = cur.ans + tmp + push[i]; 99 q.push(next); 100 } 101 } 102 } 103 } 104 return "Impossible."; 105 } 106 107 int main() 108 { 109 //freopen("poj1475.txt", "r", stdin); 110 int cnt = 0; 111 while (cin >> m >> n && m) 112 { 113 for (int i = 0; i < m; i++) 114 { 115 for (int j = 0; j < n; j++) 116 { 117 cin >> maze[i][j]; 118 if (maze[i][j] == ‘S‘) 119 { 120 pr = i; 121 pc = j; 122 } 123 else if (maze[i][j] == ‘B‘) 124 { 125 br = i; 126 bc = j; 127 } 128 else if (maze[i][j] == ‘T‘) 129 { 130 tr = i; 131 tc = j; 132 } 133 } 134 } 135 printf("Maze #%d\n", ++cnt); 136 cout << bfs_box() << endl << endl; 137 } 138 return 0; 139 }
标签:== href tmp 链接 ace include sizeof namespace span
原文地址:http://www.cnblogs.com/sench/p/7868121.html