标签：poj2991
Crane
| Time Limit: 2000MS |
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Memory Limit: 65536K |
| Total Submissions: 3777 |
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Accepted: 1031 |
|
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., 180o.
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
题意：有n根长度不尽相同的棍子，初始时它们首尾垂直相连，标号为1--n，第一根棍子的下端坐标为(0,0),上端坐标为(0,len[1]),其余棍子依次类推。接下来执行C此旋转，每次输入一个编号num和角度rad，使得第num根棍子和第num+1跟棍子间的逆时针角度变为rad度（开始因为这里WA了好几次，因为我理解成第num+1棍子相对于num顺时针旋转了rad度..），求每次旋转后第n跟棍子端点的坐标。
题解：记录第num根棍子和第num+1跟棍子间的逆时针角度pre[num],这样每次输入rad时就能确定第num+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)，它顺时针旋转A度后，得到的新向量为：x‘
= x * cos(A) + y * sin(A); y‘ = y * cos(A) - x * sin(A);
#include <stdio.h>
#include <string.h>
#include <math.h>
#define maxn 10002
#define esp 1e-9
#define lson l, mid, rt << 1
#define rson mid, r, rt << 1 | 1
const double PI = acos(-1.0);
struct Node {
double x, y, lazy;
} T[maxn << 2];
int N, C;
double pre[maxn], degree[maxn];
void build(int l, int r, int rt) {
T[rt].x = T[rt].lazy = 0.0;
if(r - l == 1) {
scanf("%lf", &T[rt].y);
return;
}
int mid = l + r >> 1;
build(lson);
build(rson);
T[rt].y = T[rt << 1].y + T[rt << 1 | 1].y;
}
void update(int L, int R, double rad, int l, int r, int rt) {
double x, y;
if(L == l && R == r) {
x = T[rt].x * cos(rad) + T[rt].y * sin(rad);
y = T[rt].y * cos(rad) - T[rt].x * sin(rad);
T[rt].x = x; T[rt].y = y;
T[rt].lazy += rad;
return;
}
int mid = l + r >> 1;
if(T[rt].lazy) {
x = T[rt<<1].x * cos(T[rt].lazy) + T[rt<<1].y * sin(T[rt].lazy);
y = T[rt<<1].y * cos(T[rt].lazy) - T[rt<<1].x * sin(T[rt].lazy);
T[rt<<1].x = x; T[rt<<1].y = y;
T[rt<<1].lazy += T[rt].lazy;
x = T[rt<<1|1].x * cos(T[rt].lazy) + T[rt<<1|1].y * sin(T[rt].lazy);
y = T[rt<<1|1].y * cos(T[rt].lazy) - T[rt<<1|1].x * sin(T[rt].lazy);
T[rt<<1|1].x = x; T[rt<<1|1].y = y;
T[rt<<1|1].lazy += T[rt].lazy;
T[rt].lazy = 0.0;
}
if(R <= mid) update(L, R, rad, lson);
else if(L >= mid) update(L, R, rad, rson);
else {
update(L, mid, rad, lson);
update(mid, R, rad, rson);
}
T[rt].y = T[rt << 1].y + T[rt << 1 | 1].y;
T[rt].x = T[rt << 1].x + T[rt << 1 | 1].x;
}
int main() {
//freopen("stdin.txt", "r", stdin);
int i, j, x, y, cas = 0;
double rad, tmp;
while(scanf("%d%d", &N, &C) == 2) {
build(0, N, 1);
if(cas++) printf("\n");
for(i = 1; i <= N; ++i)
pre[i] = PI;
while(C--) {
scanf("%d%d", &x, &y);
rad = y / 360.0 * 2 * PI;
update(x, N, pre[x] - rad, 0, N, 1);
pre[x] = rad;
printf("%.2lf %.2lf\n", T[1].x, T[1].y);
}
}
return 0;
}
POJ2991 Crane 【线段树+计算几何】
标签：poj2991
原文地址：http://blog.csdn.net/chang_mu/article/details/41510475
