HDU - 1429 胜利大逃亡(续)（BFS搜索）

bfs思路：

考虑 $BFS$搜索，考虑每个状态，所在位置 $m*n$，携带钥匙状态 $2^{10}$，总计 $20*20*1e3 = 4e5$。

代码：

#include <bits/stdc++.h>
#include <stdio.h>
#include <queue>
#include <string>
using namespace std;
const int mov[4][2] = { {1,0},{-1,0},{0,1},{0,-1} };
bool vis[25][25][(1 << 10) + 5];
struct node {
	pair<int, int> pos;
	int state, step;
	node() {
		this->step = this->state = 0;
	};
	node(int x, int y) {
		this->step = this->state = 0;
		this->pos.first = x;
		this->pos.second = y;
	};
	node(int x, int y, int state, int step) {
		this->pos.first = x;
		this->pos.second = y;
		this->state = state;
		this->step = step;
	};
};
struct Map {
	int n, m, t;
	string mp[25];
	pair<int, int> s, e;
	Map() {
		this->n = this->m = this->t = 0;
	};
};
void out(const Map& mp, const node& p) {
	printf("------------------------\nstate:");
	for (int i = 0; i < 10; i++) {
		if ((p.state & (1 << i)) != 0) {
			printf("%c ", i + 'A');
		}
	}
	printf("     step:%d\n",p.step);
	for (int i = 0; i < mp.n; i++) {
		for (int j = 0; j < mp.m; j++) {
			if (i == p.pos.first && j == p.pos.second) {
				cout << "%";
			}
			else {
				cout << mp.mp[i][j];
			}
		}
		cout << endl;
	}
}
int bfs(const Map& mp) {
	for (int i = 0; i < mp.n; i++) {
		for (int j = 0; j < mp.m; j++) {
			memset(vis[i][j], 0, sizeof(vis[i][j]));
		}
	}
	queue<node>q;
	q.push(node(mp.s.first, mp.s.second, 0, 0));	
	vis[q.front().pos.first][q.front().pos.second][q.front().state] = true;
	while (!q.empty()) {
		// out(mp, q.front());
		for (int i = 0; i < 4; i++) {
			int new_x = q.front().pos.first + mov[i][0],
				new_y = q.front().pos.second + mov[i][1],
				new_state = q.front().state,
				new_step = q.front().step + 1;
			if (new_x < 0 || new_x >= mp.n || new_y < 0 || new_y >= mp.m) { // 出界
				continue;
			}
			if (mp.mp[new_x][new_y] == '*') { // 撞墙
				continue;
			}
			if (mp.mp[new_x][new_y] <= 'j' && mp.mp[new_x][new_y] >= 'a') { // 拿钥匙
				new_state = new_state | (1 << (mp.mp[new_x][new_y] - 'a'));
			}
			if (mp.mp[new_x][new_y] <= 'J' && mp.mp[new_x][new_y] >= 'A') { // 开门
				if ((new_state & (1 << (mp.mp[new_x][new_y] - 'A'))) == 0) { // 没钥匙
					continue;
				}
			}
			if (new_x == mp.e.first && new_y == mp.e.second) { // 到达终点
				return mp.t > new_step ? new_step : -1;
			}
			if (vis[new_x][new_y][new_state] == true) { // 到达过的状态
				continue;
			}
			vis[new_x][new_y][new_state] = true;
			q.push(node(new_x, new_y, new_state, new_step));
		}
		q.pop();
	}
	return -1;
}
int main()
{
	Map x;
	while (cin >> x.n >> x.m >> x.t) {
		for (int i = 0; i < x.n; i++) {
			cin >> x.mp[i];
			for (int j = 0; j < x.m; j++) {
				if (x.mp[i][j] == '@') {
					x.s = make_pair(i, j);
				}
				if (x.mp[i][j] == '^') {
					x.e = make_pair(i, j);
				}
			}
		}
		cout << bfs(x) << endl;
	}
}