HDU - 2102 A计划 (BFS) [kuangbin带你飞]专题二

思路：接BFS判断能否在限制时间内到达公主的位置，注意如果骑士进入传送机就会被立即传送到另一层，不会能再向四周移动了，例如第一层的位置(x, y, 1)是传送机，第二层(x, y, 2)也是传送机，这种情况骑士会一直被传上传下。

AC代码： 0ms

#include<cstdio>
#include<cmath>
#include<cstring>
#include<queue>
using namespace std;
const int maxn = 15;
char G[2][maxn][maxn];
int d[2][maxn][maxn];
int n, m, Limit;

const int dx[] = {0,0,-1,1};
const int dy[] = {1,-1,0,0};
const int dz[] = {1,-1};

struct point{
	int x, y, z;
	point(){
	}
	point(int x, int y, int z):x(x), y(y), z(z){
	}
};

bool bfs() {
	memset(d, -1, sizeof(d));
	d[0][0][0] = 0;
	queue<point>q;
	q.push(point(0, 0, 0));
	while(!q.empty()) {
		point p = q.front();
		q.pop();
		int x = p.x, y = p.y, z = p.z;
		if(d[z][x][y] > Limit) return false;
		if(G[z][x][y] == 'P') return true;

		int flag = 0;
		if(G[z][x][y] == '#') { //传送机
			flag = 1;
			for(int i = 0; i < 2; ++i) {
				int pz = z + dz[i];
				if(pz < 0 || pz >= 2) continue;
				if(G[pz][x][y] == '*' || d[pz][x][y] != -1) continue;
				d[pz][x][y] = d[z][x][y];
				q.push(point(x, y, pz));
			}
		}
		if(flag) continue;

		for(int i = 0; i < 4; ++i) {
			int px = x + dx[i], py = y + dy[i];
			if(px < 0 || py < 0 || px >= n || py >= m) continue;
			if(d[z][px][py] != -1 || G[z][px][py] == '*') continue;
			d[z][px][py] = d[z][x][y] + 1;
			q.push(point(px, py, z));
		}
	}
	return false;
}
int main() {
	int T;
	scanf("%d", &T);
	while(T--) {
		scanf("%d%d%d", &n, &m, &Limit);
		for(int i = 0; i < 2; ++i) {
			for(int j = 0; j < n; ++j) scanf("%s", G[i][j]);
		}
		if(bfs()) printf("YES\n");
		else printf("NO\n");
	}
	return 0;
}

如有不当之处欢迎指出！