Gym 101981K bfs

思路：暴力让所有的骆驼和第一只骆驼合并，比如现在是第k只骆驼和第一只合并，广搜找出第k只骆驼如果想和第一只骆驼合并需要走哪一步，然后走一步，并更新所有骆驼的位置。

代码：

#include <bits/stdc++.h>
#define pii pair<int, int>
#define INF 0x3f3f3f3f
using namespace std;
const int maxn = 21;
char s[maxn][maxn];
int pre[maxn][maxn];
int dx[4] = {0, 1, 0, -1}, dy[4] = {-1, 0, 1, 0};
vector<pii> a;
vector<int> ans;
int n, m;
bool valid(pii x) {
	return x.first >= 1 && x.first <= n && x.first >= 1 && x.first <= m && s[x.first][x.second] == '1';
}
int bfs(pii st, pii ed) {
	memset(pre, -1, sizeof(pre));
	pre[st.first][st.second] = INF;
	queue<pii> q;
	q.push(st);
	while(!q.empty()) {
		pii tmp = q.front();
		q.pop();
		if(tmp == ed) {
			return pre[tmp.first][tmp.second];
		}
		for (int i = 0; i < 4; i++) {
			int x = tmp.first + dx[i], y = tmp.second + dy[i];
			if(!valid(make_pair(x, y)) || pre[x][y] != -1) continue;
			q.push(make_pair(x, y));
			pre[x][y] = (i + 2) % 4;
		}
	}
}
int main() {
	char mp[4];
	mp[0] = 'L', mp[1] = 'D', mp[2] = 'R', mp[3] = 'U';
	scanf("%d%d", &n, &m);
	for (int i = 1; i <= n; i++) {
		scanf("%s", s[i] + 1);
		for (int j = 1; j <= m; j++)
			if(s[i][j] == '1')
				a.push_back(make_pair(i, j));
	}
	int pos = 1;
	while(pos < a.size()) {
		if(a[pos] == a[0]) {
			pos++;
			continue;
		}
		while(a[pos] != a[0]) {
			int tmp = bfs(a[0], a[pos]);
			ans.push_back(tmp);
			for (int i = 0 ;i < a.size(); i++) {
				int x = a[i].first + dx[tmp], y = a[i].second + dy[tmp];
				if(valid(make_pair(x, y)))
					a[i] = make_pair(x, y);
			}
		}
	}
	for (auto x : ans) {
		printf("%c", mp[x]);
	}
}