题意：求经过所有的管道的最短路程，管道内的时间不算

思路：首先BFS处理出管道出口到其他管道入口的距离，然后在队友的指导下明白了状态转移

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
using namespace std;
const int MAXN = 16;
const int INF = 0x3f3f3f3f;

struct Node {
	int sx, sy, ex, ey;
} node[MAXN];
struct point {
	int x, y, step;
};
char g[MAXN][MAXN];
int n, m, dp[1<<MAXN][MAXN];
int dx[4]={1, -1, 0, 0};
int dy[4]={0, 0, 1 ,-1};
int vis[MAXN][MAXN], dis[MAXN][MAXN];

void bfs(int from, int to, int sx, int sy, int ex, int ey) {
	memset(vis, 0, sizeof(vis));
	queue<point> q;
	point a;
	a.x = sx, a.y = sy, a.step = 0;
	q.push(a); 
	vis[sx][sy] = 1;
	while (!q.empty()) {
		point cur = q.front();
		q.pop();
		if (cur.x == ex && cur.y == ey) {
			dis[from][to] = cur.step;
			return;
		}
		for (int i = 0; i < 4; i++) {
			int nx = cur.x + dx[i];
			int ny = cur.y + dy[i];
			if (g[nx][ny] == '#' || vis[nx][ny])
				continue;
			if (nx > 0 && nx <= n && ny > 0 && ny <= n) {
				vis[nx][ny] = 1;
				point b;
				b.x = nx, b.y = ny;
				b.step = cur.step + 1;
				q.push(b);
			}
		}
	}
}

int main() {
	while (scanf("%d%d", &n, &m) != EOF) {
		for (int i = 1; i <= n; i++)
			scanf("%s", g[i]+1);
		for (int i = 0; i < m; i++)  
			scanf("%d%d%d%d", &node[i].sx, &node[i].sy, &node[i].ex, &node[i].ey);
		memset(dis, -1, sizeof(dis));
		for (int i = 0; i < m; i++)
			for (int j = 0; j < m; j++)
				bfs(i, j, node[i].ex, node[i].ey, node[j].sx, node[j].sy);

		memset(dp, -1, sizeof(dp));
		for (int i = 0; i < m; i++)
			dp[1<<i][i] = 0;
		int all = (1<<m);
		for (int i = 0; i < all; i++) 
			for (int j = 0; j < m; j++) {
				if (i&(1<<j) && dp[i][j] != -1) {
					for (int k = 0; k < m; k++) {
						if (!(i&(1<<k)) && dis[k][j] != -1) {
							int next = i|(1<<k);
							int val = dp[i][j]+dis[k][j];
							if (dp[next][k] == -1 || val < dp[next][k])
								dp[next][k] = val;
						}
					}
				}
			}
		int ans = -1;
		for (int i = 0; i < m; i++) {
			if (dp[all-1][i] == -1)
				continue;
			if (ans == -1 || ans > dp[all-1][i])
				ans = dp[all-1][i];
		}
		printf("%d\n", ans);
	}
	return 0;
}