poj-3083 Children of the Candy Corn

#include<iostream>
#include<queue>
#include<cstring>

using namespace std;
char map[50][50];
int dl[4][2] = {{0, -1}, {-1, 0}, {0, 1}, {1, 0}};
int dr[4][2] = {{0, 1}, {-1, 0}, {0, -1}, {1, 0}};
int dx[4] = {0, 1, 0, -1};
int dy[4] = {1, 0, -1, 0};
int sx, sy, ex, ey;
int w, h;
int vis[50][50];
struct Pos{
	int x, y, s;
};

int dfs(int x, int y, int d, int step, int dir[][2]) {
	int i;
	for(i = 0; i < 4; i++) {
		int j = ((d - 1 + 4) % 4 + i) % 4;
		int nx = x + dir[j][0];
		int ny = y + dir[j][1];
		if(nx == ex && ny == ey) return step + 1;
		else if(nx < 0 || ny < 0 || nx > h || ny > w || map[nx][ny] == '#') continue;
		
		return dfs(nx, ny, j, step + 1, dir);
	}
}

int BFS(int sx, int sy) {
	queue<Pos> Q;
	Pos t = {sx, sy, 1};
	Q.push(t);
	vis[sx][sy] = 1;
	
	while(!Q.empty()) {
		Pos p = Q.front();
		Q.pop();
		if(p.x == ex && p.y == ey) return p.s;
		Pos np;
		int i;
		for(i = 0; i < 4; i++) {
			np.x = p.x + dx[i];
			np.y = p.y + dy[i];
			np.s = p.s + 1;
			if(np.x < 0 || np.y < 0 || np.x > h || np.y > w || map[np.x][np.y] == '#' || vis[np.x][np.y] == 1) continue;
			vis[np.x][np.y] = 1;
			Q.push(np);
		}
	}
	
	return -1;
}

int main() {
	int n;
	cin >> n;
	while(n--) {
		int d1, d2;
		cin >> w >> h;
		cin.get();
		int i, j;
		for(i = 0; i < h; i++) {
			for(j = 0; j < w; j++) {
				cin >> map[i][j];
				if(map[i][j] == 'S') {
					sx = i;
					sy = j;
				}
				else if(map[i][j] == 'E') {
					ex = i;
					ey = j;
				}
			}
		}
		if(sx == 0) {
			d1 = 3;
			d2 = 3;
		}
		else if(sx == h - 1) {
			d1 = 1;
			d2 = 1;
		}
		else if(sy == 0) {
			d1 = 2;
			d2 = 0;
		}
		else if(sy == w - 1) {
			d1 = 0;
			d2 = 2;
		}
		
		cout << dfs(sx, sy, d1, 1, dl) << ' ';
		cout << dfs(sx, sy, d2, 1, dr) << ' ';
		memset(vis, 0, sizeof(vis));
		cout << BFS(sx, sy) << endl;
	}
	return 0;
}

第一次做，看了题解，代码是上面这样的。

很久以后自己重新做了一遍，代码是下面这样的。。。

本题的难点就在于方向的控制吧

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
using namespace std;
int h, w;
char map[50][50];
int fol[4][2] = {0, -1, -1, 0, 0, 1, 1, 0};
int fori[4][2] = {0, 1, -1, 0, 0, -1, 1, 0};
int sx, sy, ex, ey;
int l, r, s;
struct p {
	int x, y, num;
};
void dfsl(int x, int y, int dir, int step) {
	if(l) return;
	if(x == ex && y == ey) {
		l = step + 1;
		return;
	}
	int i, j;
	for(i = 0, j = dir; i < 4; i++, j = (j + 1) % 4) {
		int tx = x + fol[j][0];
		int ty = y + fol[j][1];
		if(tx < 1 || tx > h || ty < 1 || ty > w) continue;
		if(map[tx][ty] == '#') continue;
		dfsl(tx, ty, (j + 3) % 4, step + 1);
	}
}
void dfsr(int x, int y, int dir, int step) {
	if(r) return;
	if(x == ex && y == ey) {
		r = step + 1;
		return;
	}
	int i, j;
	for(i = 0, j = dir; i < 4; i++, j = (j + 1) % 4) {
		int tx = x + fori[j][0];
		int ty = y + fori[j][1];
		if(tx < 1 || tx > h || ty < 1 || ty > w) continue;
		if(map[tx][ty] == '#') continue;
		dfsr(tx, ty, (j + 3) % 4, step + 1);
	}
}
void bfs() {
	queue<p> q;
	p t;
	map[sx][sy] = '#';
	t.x = sx;
	t.y = sy;
	t.num = 0;
	q.push(t);
	while(!q.empty()) {
		int i;
		p tt = q.front();
		q.pop();
		if(tt.x == ex && tt.y == ey) {
			s = tt.num + 1;
			return;
		}
		for(i = 0; i < 4; i++) {
			p ttt;
			ttt.x = tt.x + fol[i][0];
			ttt.y = tt.y + fol[i][1];
			if(ttt.x < 1 || ttt.x > h || ttt.y < 1 || ttt.y > w) continue;
			if(map[ttt.x][ttt.y] == '#') continue;
			map[ttt.x][ttt.y] = '#';
			ttt.num = tt.num + 1;
			q.push(ttt);
		}
	}
}
int main() {
	int T;
	scanf("%d", &T);
	while(T--) {
		scanf("%d %d", &w, &h);
		getchar();
		int i, j;
		for(i = 1; i <= h; i++) {
			for(j = 1; j <= w; j++) {
				scanf("%c", &map[i][j]);
				if(map[i][j] == 'S') sx = i, sy = j;
				else if(map[i][j] == 'E') ex = i, ey = j;
			}
			getchar();
		}
		int ordir;  //original direction
		if(sx == h) ordir = 0;
		else if(sy == 1) ordir = 1;
		else if(sx == 1) ordir = 2;
		else if(sy == w) ordir = 3; 
		l = 0;
		dfsl(sx, sy, ordir, 0);
		r = 0;
		dfsr(sx, sy, ordir, 0);
		s = 0;
		bfs();
		printf("%d %d %d\n", l, r, s);
	}
	return 0;
}