Lyft Level 5 Challenge 2018 - Final Round (Open Div. 2)

Posted 2021-02-08 dup4

 

A. The King‘s Race

签.

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 
 4 #define ll long long
 5 ll n, x, y;
 6 
 7 ll f(ll a, ll b)
 8 {
 9     return max(abs(a - x), abs(b - y));
10 }
11 
12 int main()
13 {
14     while (scanf("%lld%lld%lld", &n, &x, &y) != EOF)
15     {
16         puts(f(1, 1) <= f(n, n) ? "White" : "Black");
17     }
18     return 0;
19 }

View Code

 

 

B. Taxi drivers and Lyft

签.

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 
 4 #define N 100010
 5 int n, m;
 6 struct node
 7 {
 8     int x, t, id, small, big, small_pos, big_pos; 
 9 }a[N << 1]; 
10 int ans[N];
11 
12 int main()
13 {
14     while (scanf("%d%d", &n, &m) != EOF)
15     {
16         memset(ans, 0, sizeof ans);
17         for (int i = 1, x; i <= n + m; ++i)
18         {
19             scanf("%d", &x);
20             a[i].x = x; 
21         }
22         for (int i = 1, tmp = 0, t; i <= n + m; ++i)
23         {
24             scanf("%d", &t);
25             a[i].t = t;
26             if (t == 1)
27                 a[i].id = ++tmp;
28         }
29         int Max = -((int)1e9 + 1), pos = -1; 
30         for (int i = 1; i <= n + m; ++i)
31         {
32             if (a[i].t == 1) 
33             {
34                 Max = a[i].x;
35                 pos = a[i].id;
36             }
37             else
38             {
39                 a[i].small = Max;
40                 a[i].small_pos = pos;
41             }
42         }
43         int Min = (int)2e9 + 1; pos = -1;
44         for (int i = n + m; i >= 1; --i)
45         {
46             if (a[i].t == 1)
47             {
48                 Min = a[i].x;
49                 pos = a[i].id;
50             }
51             else
52             {
53                 a[i].big = Min;
54                 a[i].big_pos = pos;
55             }
56         }
57         for (int i = 1; i <= n + m; ++i) if (a[i].t == 0)
58         {
59             int A = abs(a[i].x - a[i].small), B = abs(a[i].x - a[i].big);
60             if (A <= B) ++ans[a[i].small_pos];
61             else ++ans[a[i].big_pos];
62         }
63         for (int i = 1; i <= m; ++i) 
64             printf("%d%c", ans[i], " 
"[i == m]);
65     }
66     return 0;
67 }

View Code

 

 

C. The Tower is Going Home

Solved.

题意：

要从$(1, 1) 走到(1e9, *)$

$有一些横着和竖着的栏杆挡着，每次只能移动到曼哈顿距离为1的格子$

$求最小的需要去掉的栏杆的数量，使得可以到达目的地$

思路：

考虑横着的栏杆，$如果左端点不是1，那么该栏杆没有用$

$那么考虑从左到右扫一遍，竖着的栏杆按顺序去掉后$

$右端点没有超过该竖着栏杆的横栏杆，该横栏杆也没有用$

更新答案即可

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 
 4 #define N 100010
 5 int n, m, q;
 6 int a[N], b[N];
 7 
 8 
 9 int main()
10 {
11     while (scanf("%d%d", &n, &m) != EOF)
12     {
13         for (int i = 1; i <= n; ++i) scanf("%d", a + i);
14         sort(a + 1, a + 1 + n);
15         q = 0;
16         for (int i = 1, x1, x2, y; i <= m; ++i)
17         {
18             scanf("%d%d%d", &x1, &x2, &y);
19             if (x1 == 1) b[++q] = x2;
20         }
21         sort(b + 1, b + 1 + q);
22         int res = q;  
23         int pos = 0;
24         a[n + 1] = (int)1e9; 
25         for (int i = 0; i <= n; ++i)
26         {
27             while (pos < q && b[pos + 1] < a[i + 1]) ++pos;
28             res = min(res, i + q - pos);
29         }
30         printf("%d
", res);
31     }
32     return 0;
33 }

View Code

 

 

D. Intersecting Subtrees

Upsolved.

题意：

有一棵树，$A选了一棵子树，B选了一棵子树$

$但是B对树的标号和A对树的标号不同$

$现在告诉你以A标号的树的形态$

$以及A, B各自选择的子树的标号$

$有5次询问机会，每次可以询问$

$A ; x;; 返回B对应的标号$

$B ; y ;; 返回A对应的标号$

最后给出答案，$A， B选择的子树是否有交，有的话输出其中一个交点，否则输出-1$

思路：

随便选一个$B中的点，得到A中的点，如果刚好是A子树内的，直接输出$

$否则以这个点去找一个最近的点，再询问一次，如果是就输出，否则就是-1$

$因为找到的是最近的属于A选择子树内的点，说明这条路径上没有交，如果这个点不是交$

$那么说明那个点那头也不会有交$ 

$如果有交的话，这个点就会把B选择的子树分成两部分，就不是连通的，和题意想违背$

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 
 4 #define N 1010
 5 int t, n, k, st;
 6 int visa[N], visb[N];
 7 vector <int> G[N];
 8 
 9 int fa[N];
10 void BFS(int S) 
11 {
12     queue <int> q; q.push(S); fa[st] = S;
13     while (!q.empty())
14     {
15         int u = q.front(); q.pop();
16         if (visa[u]) 
17         {
18             st = u;
19             return; 
20         }
21         for (auto v : G[u]) if (v != fa[u])
22         {
23             fa[v] = u;
24             q.push(v);
25         }
26     }
27 }
28 
29 int dfs(int num,int fa){
30     if(visa[num]){
31         return num;
32     }
33     for (auto v : G[num]) if (v != fa) 
34     {
35         int temp=dfs(v,num);
36         if(temp!=-1){
37             return temp;
38         }
39     }
40     return -1;
41 }
42 
43 int main()
44 {
45     scanf("%d", &t);
46     while (t--)
47     {
48         scanf("%d", &n);
49         for (int i = 1; i <= n; ++i) G[i].clear();
50         memset(visa, 0, sizeof visa);
51         memset(visb, 0, sizeof visb);
52         for (int i = 1, u, v; i < n; ++i)
53         {
54             scanf("%d%d", &u, &v);
55             G[u].push_back(v);
56             G[v].push_back(u);
57         }
58         scanf("%d", &k);
59         for (int i = 1, x; i <= k; ++i)
60         {
61             scanf("%d", &x);
62             visa[x] = 1;
63         }    
64         scanf("%d", &k);
65         for (int i = 1, x; i <= k; ++i)
66         {
67             scanf("%d", &x);
68             visb[x] = 1;
69             st = x;
70         }
71         printf("B %d
", st);
72         fflush(stdout);
73         st = -1;
74         scanf("%d", &st); 
75         if (visa[st])
76         {
77             printf("C %d
", st);
78             fflush(stdout);
79             continue;
80         }
81         BFS(st);
82         //st = dfs(st, st);
83         int ed;
84         printf("A %d
", st);
85         fflush(stdout);
86         scanf("%d", &ed);
87         if (visa[st] && visb[ed])
88             printf("C %d
", st);
89         else 
90             puts("C -1");
91         fflush(stdout);
92     }
93     return 0;
94 }

View Code

 

 

E. Optimal Polygon Perimeter

Upsolved.

题意：

给出一个凸包，两点之间的距离为曼哈顿距离

定义$f(x) 为选择x个点构成一个凸包的最大周长$

输出$f(3), f(4) cdots f(n)$

思路：

对于$n >= 4的答案，就是选择最大上界，最大下界，最大左界，最大右界，构成的矩形的周长$

$再考虑n == 3的时候$

$因为是一个三角形，那么肯定是某两个最构成的两个点，再加上一个点，那个点枚举求解即可$

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 
 4 #define N 300010
 5 int n;
 6 int x[N], y[N];
 7 
 8 int main()
 9 {
10     while (scanf("%d", &n) != EOF)
11     {
12         int Max[2], Min[2];
13         Min[0] = Min[1] = (int)1e9;
14         Max[0] = Max[1] = -(int)1e9;
15         for (int i = 1; i <= n; ++i)
16         {
17             scanf("%d%d", x + i, y + i);
18             Max[0] = max(Max[0], x[i]);
19             Min[0] = min(Min[0], x[i]);
20             Max[1] = max(Max[1], y[i]);
21             Min[1] = min(Min[1], y[i]);
22         }
23         if (n == 3)
24             printf("%d
", 2 * (Max[0] + Max[1] - Min[0] - Min[1]));
25         else
26         {
27             int res = 0;
28             for (int i = 1; i <= n; ++i)
29                 res = max(res, max(abs(x[i] - Max[0]), abs(x[i] - Min[0])) + max(abs(y[i] - Max[1]), abs(y[i] - Min[1])));
30             res *= 2;
31             printf("%d", res);    
32             for (int i = 4; i <= n; ++i) 
33                 printf(" %d", 2 * (Max[0] + Max[1] - Min[0] - Min[1]));
34             puts("");
35         }
36     }
37     return 0;
38 }

View Code