POJ 2991 Crane（线段树）

Crane

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 7687		Accepted: 2075		Special Judge

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

2 1
10 5
1 90

3 2
5 5 5
1 270
2 90

Sample Output

5.00 10.00

-10.00 5.00
-5.00 10.00

Source

CTU Open 2005

 

节点i表示的向量是vxi，vyi，角度是angi，两个儿子节点是chl和chr，

 

vx[i]=vx[chl]+ (cos(angi)*vx[chr]-sin(angi)*vy[chr]);
vy[i]=vy[chl]+ (sin(angi)*vx[chr]+cos(angi)*vy[chr]);

 

#include <iostream>  
#include <cmath>  
#include <cstdio>  
#define MAX_N 10000  
#define M_PI 3.14159265358979323846  
using namespace std;  
  
const int ST_SIZE=(1<<15)-1;  
//输入  
int N,C;  
int L[MAX_N+5];  
int S[MAX_N+5],A[MAX_N+5];  
  
//线段树所维护的数据  
double vx[ST_SIZE],vy[ST_SIZE];  
double ang[ST_SIZE];  
  
//为了查询角度的变化而保存的当前角度的数组  
double prv[MAX_N];  
  
//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;  
    else 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 s=sin(ang[v]),c=cos(ang[v]);     //围绕原点的旋转：  
        vx[v]=vx[chl]+(c*vx[chr]-s*vy[chr]);    //x' = x * cos(a) - y * sin(a)  
        vy[v]=vy[chl]+(s*vx[chr]+c*vy[chr]);    //y' = x * sin(a) + y * cos(a)  
    }  
  
}  
void solve()  
{  
    //初始化  
    init(0,0,N);  
    for(int i=1;i<N;i++)  
        prv[i]=M_PI;  
    //处理操作  
    for(int i=0;i<C;i++)  
    {  
        int s=S[i];  
        double a=A[i]/360.0*2*M_PI;     //把角度换算为弧度  
        change(s,a-prv[s],0,0,N);  
        prv[s]=a;  
        printf("%.2f %.2f\n",vx[0],vy[0]);  
  
    }  
  
  
}  
int main()  
{  
    while(cin>>N>>C)  
    {  
        for(int i=0;i<N;i++)  
            scanf("%d",&L[i]);  
        for(int i=0;i<C;i++)  
            scanf("%d%d",&S[i],&A[i]);  
        solve();  
  
    }  
  
    return 0;  
}