POJ 2991 Crane 线段树

这个题比较神奇。大意就是有一个长杆，杆上有一些关节点，可以转动，然后每次操作就是将某个关节点的角度变成某个角度，最后求末尾的坐标。

线段树每个结点存的信息是，这一段末尾的x坐标，y坐标，以及该段与y轴的夹角，其中x,y坐标是与上一段线段的相对坐标。这样的好处是显而易见的。注意，这个与y轴的夹角是该段线段顺时针旋转至初始状态的角度，如下图

然后每次更新的时候需要注意，是把某个结点后的所有线段都旋转

/*ID: sdj22251PROG: inflateLANG: C++*/#include <iostream>#include <vector>#include <list>#include <map>#include <set>#include <deque>#include <queue>#include <stack>#include <bitset>#include <algorithm>#include <functional>#include <numeric>#include <utility>#include <sstream>#include <iomanip>#include <cstdio>#include <cmath>#include <cstdlib>#include <cctype>#include <string>#include <cstring>#include <cmath>#include <ctime>#define MAXN 10005#define INF 1000000000#define L(x) x<<1#define R(x) x<<1|1#define PI acos(-1.0)#define eps 1e-7using namespace std;int len[MAXN];struct node{    int left, right, mid;    double x, y;    int turn;}tree[4 * MAXN];void up(int C){    tree[C].y = tree[L(C)].y + tree[R(C)].y;    tree[C].x = tree[L(C)].x + tree[R(C)].x;}void change(int C, int turn){    tree[C].turn = (tree[C].turn + turn) % 360;    double tx = tree[C].x;    double ty = tree[C].y;    double degree = turn * 1.0 / 180.0 * PI;    tree[C].x = tx * cos(degree) - ty * sin(degree);    tree[C].y = tx * sin(degree) + ty * cos(degree);}void down(int C){    if(tree[C].turn)    {        change(L(C), tree[C].turn);        change(R(C), tree[C].turn);        tree[C].turn = 0;    }}void make_tree(int s, int e, int C){    tree[C].left = s;    tree[C].right = e;    tree[C].mid = (s + e) >> 1;    tree[C].x = 0;    tree[C].turn = 0;    if(s == e) {tree[C].y = len[s]; return;}    make_tree(s, tree[C].mid, L(C));    make_tree(tree[C].mid + 1, e, R(C));    up(C);}int query(int C, int p){    if(tree[C].left == tree[C].right)        return tree[C].turn;    down(C);    if(tree[C].mid >= p) return query(L(C), p);    else return query(R(C), p);}void update(int s, int e, int turn, int C){    if(tree[C].left >= s && tree[C].right <= e)    {        change(C, turn);        return;    }    down(C);    if(tree[C].mid >= s) update(s, e, turn, L(C));    if(tree[C].mid < e) update(s, e, turn, R(C));    up(C);}int main(){    int n, c;    int x, y;    bool fg = 0;    while(scanf("%d%d", &n, &c) != EOF)    {        if(fg) puts("");        fg = 1;        for(int i = 1; i <= n; i++)            scanf("%d", &len[i]);        make_tree(1, n, 1);        for(int i = 1; i <= c; i++)        {            scanf("%d%d", &x, &y);            int pos1 = query(1, x);            int pos2 = query(1, x + 1);            int t = (-180 + pos1 - pos2 + y + 360) % 360;            update(x + 1, n, t, 1);            printf("%.2f %.2f\n", tree[1].x + eps, tree[1].y + eps);        }    }    return 0;}