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], " \n"[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\n", 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\n", st);
72 fflush(stdout);
73 st = -1;
74 scanf("%d", &st);
75 if (visa[st])
76 {
77 printf("C %d\n", st);
78 fflush(stdout);
79 continue;
80 }
81 BFS(st);
82 //st = dfs(st, st);
83 int ed;
84 printf("A %d\n", st);
85 fflush(stdout);
86 scanf("%d", &ed);
87 if (visa[st] && visb[ed])
88 printf("C %d\n", 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\n", 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