poj2991 Crane

Posted 2020-10-26 王宜鸣

思路：

线段树每个节点维护第一条线段起点指向最后一条线段终点的向量，于是每一个操作都是一次区间更新。使用成段更新的线段树即可。
实现：

 1 #include <cstdio>
 2 #include <iostream>
 3 #include <cmath>
 4 #include <cstring>
 5 
 6 using namespace std;
 7 
 8 const int MAXN = 10005;
 9 const double PI = acos(-1.0);
10 
11 int a[MAXN], d[MAXN], n, c;
12 struct node
13 {
14     double x, y;
15     int lazy;
16 };
17 node tree[MAXN << 2];
18 double trans(int d) { return (double)d * PI / 180.0; }
19 void rotate(double & x, double & y, int dx)
20 {
21     double tx = x * cos(trans(dx)) - y * sin(trans(dx));
22     double ty = y * cos(trans(dx)) + x * sin(trans(dx));
23     x = tx; y = ty;
24 }
25 void pushdown(int num)
26 {
27     if (!tree[num].lazy) return;
28     int tmp = tree[num].lazy;
29     rotate(tree[num << 1].x, tree[num << 1].y, tmp);
30     rotate(tree[num << 1 | 1].x, tree[num << 1 | 1].y, tmp);
31     tree[num << 1].lazy += tmp;
32     tree[num << 1 | 1].lazy += tmp;
33     tree[num].lazy = 0;
34 }
35 void pushup(int num)
36 {
37     tree[num].x = tree[num << 1].x + tree[num << 1 | 1].x;
38     tree[num].y = tree[num << 1].y + tree[num << 1 | 1].y;
39 }
40 void build(int num, int l, int r)
41 {
42     if (l == r) { tree[num].y = a[r]; return; }
43     int m = l + r >> 1;
44     build(num << 1, l, m);
45     build(num << 1 | 1, m + 1, r);
46     pushup(num);
47 }
48 void update(int num, int l, int r, int x, int y, int dx)
49 {
50     if (x <= l && y >= r) 
51     { 
52         rotate(tree[num].x, tree[num].y, dx);
53         tree[num].lazy += dx;
54         return;
55     }
56     int m = l + r >> 1;
57     pushdown(num);
58     if (x <= m) update(num << 1, l, m, x, y, dx);
59     if (y >= m + 1) update(num << 1 | 1, m + 1, r, x, y, dx);
60     pushup(num);
61 }
62 
63 int main()
64 {
65     while (scanf("%d %d", &n, &c) != EOF)
66     {
67         for (int i = 1; i <= n * 4; i++) { tree[i].x = tree[i].y = 0; tree[i].lazy = 0; }
68         for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
69         for (int i = 1; i < n; i++) d[i] = 180;
70         build(1, 1, n);
71         int x, y;
72         for (int i = 0; i < c; i++)
73         {
74             scanf("%d %d", &x, &y);
75             int dx = y - d[x];
76             d[x] = y;
77             update(1, 1, n, x + 1, n, dx);
78             printf("%.2f %.2f\n", tree[1].x, tree[1].y);
79         }
80         puts("");
81     }
82     return 0;
83 }