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链接：https://codeforces.com/problemset/problem/173/B

题目：

思路：

初识01BFS

什么是 01 BFS 呢？通常的 BFS 为一步权值为 1，而某些题需要的不是走到步数，而是某种操作数，如花费一个操作可以走 k 步，而不花费只能走 1 步，通常我们使用双端队列可插队的性质来进行代码的编写，具体的对于不花费，那么就插入到前面，而对于花费则插入到最后

本题中操作为 “四射”，所以按照上面描述很快能写出来代码

代码：

#include <iostream>
#include <algorithm>
#include<cstring>
#include<cctype>
#include<string>
#include <set>
#include <vector>
#include <cmath>
#include <queue>
#include <unordered_set>
#include <map>
#include <unordered_map>
#include <stack>
#include <memory>
#include <complex>
using namespace std;
#define int long long
#define yes cout << "Yes\n"
#define no cout << "No\n"
pair<int, int> dir[4] = { {-1,0},{0,1},{1,0},{0,-1} };
int use[1005][1005];
int dis[1005][1005][4];
deque<int> q;void af(int x,int y,int d,int ans)
{if (ans < dis[y][x][d]){dis[y][x][d] = ans;q.push_front(d);q.push_front(y);q.push_front(x);}
}
void ab(int x, int y, int d, int ans)
{if (ans < dis[y][x][d]){dis[y][x][d] = ans;q.push_back(x);q.push_back(y);q.push_back(d);}
}
void solve()
{int n, m;cin >> n >> m;vector<string> mp(n);for (int i = 0; i < n; i++){cin >> mp[i];}for (int i = 0; i < n; i++){for (int j = 0; j < m; j++){for (int k = 0; k < 4; k++){dis[i][j][k] = 1e18;}}}af(m - 1, n - 1, 0, 0);while (!q.empty()){int x = q[0];int y = q[1];int d = q[2];q.pop_front();q.pop_front();q.pop_front();int ans = dis[y][x][d];int nx = x + dir[d].first;int ny = y + dir[d].second;if (nx >= 0 && nx < m && ny >=0 && ny < n){af(nx, ny, d, ans);}if (mp[y][x] == '#'){for (int i = 0; i < 4; i++){if (i != d)ab(x, y, i, ans + 1);}}}if (dis[0][0][0] != 1e18)cout << dis[0][0][0] << endl;elsecout << "-1\n";
}signed main()
{ios::sync_with_stdio(false);cin.tie(nullptr);int t = 1;while (t--){solve();}return 0;
}