POJ 2991 Crane

Posted 2020-09-27 EricJeffrey

题目链接：http://poj.org/problem?id=2991

挑战程序设计竞赛书上的例题。书上的思路还是蛮巧妙的，第一遍看真的没怎么看太懂，今天算是终于弄懂了。

稍微思考即可想到如果都用向量表示的话实际上就是从s+1到n的区间更新了，每个向量都旋转一定角度最后终点坐标实际上是全部向量的和。因此可以使用线段树进行区间更新。

书上更新的时候实际上是----对于包含了s的区间更新坐标，不包含s的区间只设置一个标志表示这一段区间已经旋转过了一定角度，每次需要更新的时候直接用这个角度计算即可。

另一方面就是旋转后坐标计算公式，没啥难度。。。。

代码：

  1 #include <map>
  2 #include <set>
  3 #include <cmath>
  4 #include <queue>
  5 #include <stack>
  6 #include <cstdio>
  7 #include <string>
  8 #include <vector>
  9 #include <cstdlib>
 10 #include <cstring>
 11 #include <sstream>
 12 #include <iostream>
 13 #include <algorithm>
 14 #include <functional>
 15 using namespace std;
 16 #define rep(i,a,n) for (int i=a;i<n;i++)
 17 #define per(i,a,n) for (int i=n-1;i>=a;i--)
 18 #define all(x) (x).begin(),(x).end()
 19 #define pb push_back
 20 #define mp make_pair
 21 #define lson l,m,rt<<1
 22 #define rson m+1,r,rt<<1|1
 23 typedef long long ll;
 24 typedef vector<int> VI;
 25 typedef pair<int, int> PII;
 26 const ll MOD = 1e9 + 7;
 27 const int inf = 0x3f3f3f3f;
 28 #define M_PI 3.14159265358979323846
 29 //head
 30 #define maxn 70000
 31 struct node{
 32     int left, right;
 33     double vx, vy;
 34     double angle;
 35     node() {}
 36     node(int l, int r, double x, double y, double ang): left(l), right(r), vx(x), vy(y), angle(ang){}
 37 };
 38 node tree[maxn];
 39 double prv[maxn];
 40 int n, c;
 41 int len[maxn];
 42 
 43 void build(int l, int r, int root){
 44     tree[root] = node(l, r, 0, 0, 0);
 45     if(l == r){
 46         tree[root].vy = len[l];
 47         return;
 48     }
 49     int mid = (l + r) / 2;
 50     build(l, mid, root << 1);
 51     build(mid + 1, r, root << 1 | 1);
 52     tree[root].vy = tree[root << 1].vy + tree[root << 1 | 1].vy;
 53     return;
 54 } 
 55 //rotate seg s+1 but won‘t change the coordinate of them 
 56 void update(int s, double a, int l, int r, int k){
 57     if(s < l)    
 58         return;
 59     if(s == r && l == r)
 60         return;
 61     else if(s <= r){
 62         int lc = k << 1, rc = k << 1 | 1;
 63         int mid = (l + r) / 2;
 64         update(s, a, l, mid, lc);
 65         update(s, a, mid + 1, r, rc);
 66         
 67         if(s <= mid) 
 68             tree[k].angle += a;
 69         double tmang = tree[k].angle;
 70         tree[k].vx = tree[lc].vx + (cos(tmang) * tree[rc].vx - sin(tmang) * tree[rc].vy);
 71         tree[k].vy = tree[lc].vy + (sin(tmang) * tree[rc].vx + cos(tmang) * tree[rc].vy);
 72     }
 73 }
 74 
 75 int main(){
 76     while(scanf("%d %d", &n, &c) != EOF){
 77         memset(len, 0, sizeof(len));
 78         memset(tree, 0, sizeof(tree));
 79         for(int i = 0; i < n; i++)
 80             prv[i] = M_PI;
 81         for(int i = 0; i < n; i++)
 82             scanf("%d", &len[i]);
 83         build(0, n - 1, 1);
 84         
 85         while(c--){
 86             int s, a;
 87             scanf("%d %d", &s, &a);
 88             s--;
 89             double tma = a / 180.0 * M_PI;
 90             update(s, tma - prv[s] , 0, n - 1, 1);
 91             prv[s] = tma;
 92              97             printf("%.2f %.2f\n", ansx, ansy);
 98         }
 99         puts("");
100     }
101 }

题目：

Crane

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 6807		Accepted: 1822		Special Judge

Description

ACM has bought a new crane (crane -- je?áb) . 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