poj2991(Crane)线段树+计算几何

Description

ACM has bought a new crane (crane -- jeřáb) . The crane consists of n segments of various lengths, connected by flexible joints. The end of the i-th segment is joined to the beginning of the i + 1-th one, for 1 ≤ i < n. The beginning of the first segment is fixed at point with coordinates (0, 0) and its end at point with coordinates (0, w), where w is the length of the first segment. All of the segments lie always in one plane, and the joints allow arbitrary rotation in that plane. After series of unpleasant accidents, it was decided that software that controls the crane must contain a piece of code that constantly checks the position of the end of crane, and stops the crane if a collision should happen.

Your task is to write a part of this software that determines the position of the end of the n-th segment after each command. The state of the crane is determined by the angles between consecutive segments. Initially, all of the angles are straight, i.e., 180 ^o. The operator issues commands that change the angle in exactly one joint.

Input

The input consists of several instances, separated by single empty lines.

The first line of each instance consists of two integers 1 ≤ n ≤10 000 and c 0 separated by a single space -- the number of segments of the crane and the number of commands. The second line consists of n integers l1,..., ln (1 li 100) separated by single spaces. The length of the i-th segment of the crane is li. The following c lines specify the commands of the operator. Each line describing the command consists of two integers s and a (1 ≤ s < n, 0 ≤ a ≤ 359) separated by a single space -- the order to change the angle between the s-th and the s + 1-th segment to a degrees (the angle is measured counterclockwise from the s-th to the s + 1-th segment).

Output

The output for each instance consists of c lines. The i-th of the lines consists of two rational numbers x and y separated by a single space -- the coordinates of the end of the n-th segment after the i-th command, rounded to two digits after the decimal point.

The outputs for each two consecutive instances must be separated by a single empty line.

Sample Input

Sample Output

5.00 10.00-10.00 5.00
-5.00 10.00

题意：n根相连的木棍初始时竖直摆放，从下到上编号依次为1~n。给出每根木棍长度和c次操作。每次操作两个操作数s和a，表示转动s+1，使得s和s+1夹角变为a，其中夹角指的是s逆时针转到s+1转过的角度。输出每次操作之后最后一根木棍尾端点坐标，其中原点为第一根木棍头端点。

题解：记录第s根棍子到第s+1根棍子间的逆时针角度prv[s],这样每次输入a时就能确定第s+1根及以后棍子的绝对旋转角度，然后用这个角度维护区间值。对于一个向量(x, y)，它逆时针旋转A度后，得到的新向量为：x' = xcos(A) - ysin(A)，y' = xsin(A) + ycos(A)。

#include <iostream>
#include <algorithm>
#include <cstdio>
using namespace std;const double PI = acos(-1.0);
const int st_size = (1 << 15) - 1;
const int maxn = 10010;//输入
int n, c;
int L[maxn];//线段树维护的数据
double vx[st_size], vy[st_size];    //各节点的向量
double ang[st_size];    //各节点的角度double prv[maxn];//初始化线段树
//k是节点的编号，l, r表示当前节点对应的是[l, r]区间
void init(int k, int l, int r)
{ang[k] = vx[k] = 0.0;if (r - l == 1)     //叶子节点vy[k] = L[l];else    //非叶子节点{int chl = k * 2 + 1, chr = k * 2 + 2;init(chl, l, (l+r)/2);init(chr, (l+r)/2, r);vy[k] = vy[chl] + vy[chr];}
}//把s和s+1的角度变为a
//v是节点的编号，l, r表示当前节点对应的是[l, r)区间
void change(int s, double a, int v, int l, int r)
{if (s <= l)return;if (s < r){int chl = v * 2 + 1, chr = v * 2 + 2;int m = (l + r) / 2;change(s, a, chl, l, m);change(s, a, chr, m, r);if (s <= m)ang[v] += a;double si = sin(ang[v]), co = cos(ang[v]);vx[v] = vx[chl] + co * vx[chr] - si * vy[chr];vy[v] = vy[chl] + si * vx[chr] + co * vy[chr];}
}int main()
{int cnt = 0;while (~scanf("%d%d", &n, &c)){if (cnt++)printf("\n");for (int i = 0; i < n; ++i)scanf("%d", &L[i]);//初始化init(0, 0, n);fill(prv+1, prv+n, PI);while (c--){int s;double a;scanf("%d%lf", &s, &a);a = a / 180.0 * PI;   //把角度换算成弧度change(s, a-prv[s], 0, 0, n);prv[s] = a;printf("%.2f %.2f\n", vx[0], vy[0]);}}return 0;
}

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