UVA 589 - Pushing Boxes（BFS+状态判重）

题目链接：589 - Pushing Boxes

思路：广搜+状态判重，用人的位置和箱子位置和当前步数作为状态。然后由于是要优先推箱子次数少，所以利用优先队列去取状态。

代码：

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <vector>
#include <queue>
#include <string>
#include <map>
#define INF 0x3f3f3f3f
using namespace std;
const int d[4][2] = {{-1, 0}, {1, 0}, {0, 1}, {0, -1}};
const char s1[10] = "NSEW";
const char s2[10] = "nsew";
const int N = 25;
int r, c, push_num;
char g[N][N];
string ans;
map<int, bool> vis[20][20][20][20];
typedef pair<int, int> pi;
pi s, b, t;

struct State {
	int sx, sy, bx, by, num;
	string path;
	State() {}
	State(int _sx, int _sy, int _bx, int _by, int _num) {
		sx = _sx; sy = _sy; bx = _bx; by = _by; num = _num;
		path = "";
	}
	friend bool operator < (State a, State b) {
		if (a.num != b.num)
			return a.num > b.num;
		return a.path.length() > b.path.length();
	}
};

void init() {
	ans = "";
	push_num = INF;
	for (int i = 0; i < r; i++) {
		scanf("%s", g[i]);
		for (int j = 0; j < c; j++) {
			if (g[i][j] == ‘S‘)
				s = make_pair(i, j);
			else if (g[i][j] == ‘B‘)
				b = make_pair(i, j);
			else if (g[i][j] == ‘T‘)
				t = make_pair(i, j);
			if (g[i][j] != ‘#‘)
				g[i][j] = ‘.‘;
			for (int k = 0; k < r; k++)
				for (int l = 0; l < c; l++)
					vis[i][j][k][l].clear();
		}
	}
}

bool check(int x, int y) {
	if (x < 0 || x >= r || y < 0 || y >= c || g[x][y] != ‘.‘) return false;
	return true;
}

void solve() {
	priority_queue<State> Q;
	vis[s.first][s.second][b.first][b.second][0] = 1;
	Q.push(State(s.first, s.second, b.first, b.second, 0));
	while (!Q.empty()) {
		State p = Q.top();
		Q.pop();
		if (p.bx == t.first && p.by == t.second) {
			if (push_num > p.num) {
				ans = p.path;
				push_num = p.num;
			} else if (push_num == p.num) {
				if (ans.length() > p.path.length())
					ans = p.path;
			}
			else break;
			continue;
		}
		for (int i = 0; i < 4; i++) {
			State q = p;
			q.sx += d[i][0];
			q.sy += d[i][1];
			if (!check(q.sx, q.sy)) continue; 
			q.path += s2[i];
			if (q.sx == q.bx && q.sy == q.by) {
				q.bx += d[i][0], q.by += d[i][1];
				if (!check(q.bx, q.by)) continue;
				q.num++;
				if (q.num > push_num) continue;
				q.path[q.path.length() - 1] = s1[i];
			}
			if (vis[q.sx][q.sy][q.bx][q.by][ans.length()]) continue;
			vis[q.sx][q.sy][q.bx][q.by][ans.length()] = true;
			Q.push(q);
		}
	}
}

int main() {
	int cas = 0;
	while (~scanf("%d%d", &r, &c) && r || c) {
		init();
		printf("Maze #%d\n", ++cas);
		solve();
		if (ans.length())
			cout << ans << endl;
		else printf("Impossible.\n");
		printf("\n");
	}
	return 0;
}

UVA 589 - Pushing Boxes（BFS+状态判重）,布布扣,bubuko.com

UVA 589 - Pushing Boxes（BFS+状态判重）

原文：http://blog.csdn.net/accelerator_/article/details/22727359