广搜套路题，就是在求最短路的基础上多了一个查克拉的限制，我们定义一个结构体，里面存储到了每个点的坐标，步数，以及查克拉，然后去周围点的时候对正常的通路和大蛇丸手下，做不同处理即可。第一次走到终点，直接输出步数即可。

#include <bits/stdc++.h>
#define ll long long
#define inf 0x3f3f3f3f
#define pb push_back
#define fi first
#define se second
#define re register
using namespace std;
typedef pair<ll, ll> pi;
int m, n, t, sx, sy, ex, ey;
char a[205][205];
int dx[4] = {-1, 0, 1, 0}, dy[4] = {0, 1, 0, -1};
struct node
{
    int x, y, time, hp;
};
bool vis[205][205], flag;
queue<node> q;
void bfs()
{
    q.push({sx, sy, 0, t});
    vis[sx][sy] = 1;
    while (q.size())
    {
        node now = q.front();
        q.pop();
        if (now.x == ex && now.y == ey)
        {
            flag = true;
            cout << now.time;
            return ;
        }
        for (int i = 0; i < 4; i++)
        {
            int nx = now.x + dx[i], ny = now.y + dy[i];
            if (nx < 1 || nx > m || ny < 1 || ny > n) continue;
            if (!vis[nx][ny])
            {
                vis[nx][ny] = 1;
                if (a[nx][ny] == '*' || a[nx][ny] == '+') q.push({nx, ny, now.time + 1, now.hp});
                else if (a[nx][ny] == '#' && now.hp > 0) q.push({nx, ny, now.time + 1, now.hp - 1}); 
            }
        }
    }
}
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cin >> m >> n >> t;
    for (int i = 1; i <= m; i++)
    {
        for (int j = 1; j <= n; j++)
        {
            cin >> a[i][j];
            if (a[i][j] == '@') sx = i, sy = j;
            if (a[i][j] == '+') ex = i, ey = j;
        }
    }
    bfs();
    if (!flag) cout << -1;
    return 0;
}