poj2991 Crane

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：

 #include <cstdio>
 #include <iostream>
 #include <cmath>
 #include <cstring>
 
 using namespace std;
 
 const int MAXN = ;
 const double PI = acos(-1.0);
 
 int a[MAXN], d[MAXN], n, c;
 struct node
 {
     double x, y;
     int lazy;
 };
 node tree[MAXN << ];
 double trans(int d) { return (double)d * PI / 180.0; }
 void rotate(double & x, double & y, int dx)
 {
     double tx = x * cos(trans(dx)) - y * sin(trans(dx));
     double ty = y * cos(trans(dx)) + x * sin(trans(dx));
     x = tx; y = ty;
 }
 void pushdown(int num)
 {
     if (!tree[num].lazy) return;
     int tmp = tree[num].lazy;
     rotate(tree[num << ].x, tree[num << ].y, tmp);
     rotate(tree[num <<  | ].x, tree[num <<  | ].y, tmp);
     tree[num << ].lazy += tmp;
     tree[num <<  | ].lazy += tmp;
     tree[num].lazy = ;
 }
 void pushup(int num)
 {
     tree[num].x = tree[num << ].x + tree[num <<  | ].x;
     tree[num].y = tree[num << ].y + tree[num <<  | ].y;
 }
 void build(int num, int l, int r)
 {
     if (l == r) { tree[num].y = a[r]; return; }
     int m = l + r >> ;
     build(num << , l, m);
     build(num <<  | , m + , r);
     pushup(num);
 }
 void update(int num, int l, int r, int x, int y, int dx)
 {
     if (x <= l && y >= r)
     {
         rotate(tree[num].x, tree[num].y, dx);
         tree[num].lazy += dx;
         return;
     }
     int m = l + r >> ;
     pushdown(num);
     if (x <= m) update(num << , l, m, x, y, dx);
     if (y >= m + ) update(num <<  | , m + , r, x, y, dx);
     pushup(num);
 }
 
 int main()
 {
     while (scanf("%d %d", &n, &c) != EOF)
     {
         for (int i = ; i <= n * ; i++) { tree[i].x = tree[i].y = ; tree[i].lazy = ; }
         for (int i = ; i <= n; i++) scanf("%d", &a[i]);
         for (int i = ; i < n; i++) d[i] = ;
         build(, , n);
         int x, y;
         for (int i = ; i < c; i++)
         {
             scanf("%d %d", &x, &y);
             int dx = y - d[x];
             d[x] = y;
             update(, , n, x + , n, dx);
             printf("%.2f %.2f\n", tree[].x, tree[].y);
         }
         puts("");
     }
     return ;
 }