hdu1732 Push Box (bfs)

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三个箱子的推箱子。

节点保存人的坐标，三个箱子的坐标和整个地图。

判重用一个8维bool数组，表示人的坐标和三个箱子的坐标。

注意当箱子的前方还有箱子的时候不能前进。

代码：

#include <cstdio>#include <cstring>#include <algorithm>#include <iostream>#include <queue>using namespace std;struct Node{int x, y;int step;char map[9][9];int box[3][2];Node() {x = y = step = 0;memset(map, 0, sizeof(map));memset(box, 0, sizeof(box));}}s, cur, nex;int n, m;int dir[4][2] = {{0, -1}, {0, 1}, {-1, 0}, {1, 0}};bool vis[9][9][9][9][9][9][9][9];bool done() {return s.map[cur.box[0][0]][cur.box[0][1]] == '@'&& s.map[cur.box[1][0]][cur.box[1][1]] == '@'&& s.map[cur.box[2][0]][cur.box[2][1]] == '@';}bool check(int x, int y) {return (x >= 0 && x < n && y >= 0 && y < m && s.map[x][y] != '#');}int bfs() {memset(vis, 0, sizeof(vis));queue<Node> q;q.push(s);vis[s.x][s.y][s.box[0][0]][s.box[0][1]][s.box[1][0]][s.box[1][1]][s.box[2][0]][s.box[2][1]] = 1;while (!q.empty()) {cur = q.front();q.pop();if (done()) return cur.step;for (int i = 0; i < 4; i++) {nex = cur;int nx = cur.x + dir[i][0];int ny = cur.y + dir[i][1];if (!check(nx, ny)) continue;if (vis[nx][ny][nex.box[0][0]][nex.box[0][1]][nex.box[1][0]][nex.box[1][1]][nex.box[2][0]][nex.box[2][1]])continue;if (nex.map[nx][ny] == '*') {int bx = nx + dir[i][0];int by = ny + dir[i][1];if ((!check(bx, by)) || nex.map[bx][by] == '*') continue;int box;for (box = 0; box < 3; box++)if (nex.box[box][0] == nx && nex.box[box][1] == ny)break;nex.map[bx][by] = '*';nex.map[nx][ny] = nex.map[cur.x][cur.y] = '.';if (s.map[cur.x][cur.y] == '@') nex.map[cur.x][cur.y] = '@';if (s.map[nx][ny] == '@') nex.map[nx][ny] = '@';nex.box[box][0] = bx;nex.box[box][1] = by;if ((vis[nx][ny][nex.box[0][0]][nex.box[0][1]][nex.box[1][0]][nex.box[1][1]][nex.box[2][0]][nex.box[2][1]]))continue;}nex.step++;nex.x = nx, nex.y = ny;vis[nx][ny][nex.box[0][0]][nex.box[0][1]][nex.box[1][0]][nex.box[1][1]][nex.box[2][0]][nex.box[2][1]] = 1;q.push(nex);}}return -1;}int main() {while (~scanf("%d %d", &n, &m)) {s = Node();for (int i = 0; i < n; i++)scanf("%s", s.map[i]);int t = 0;for (int i = 0; i < n; i++) {for (int j = 0; j < m; j++) {if (s.map[i][j] == 'X') {s.x = i;s.y = j;}if (s.map[i][j] == '*') {s.box[t][0] = i;s.box[t++][1] = j;}}}int ans = bfs();cout << ans << endl;}return 0;}

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