洛谷P2903 [USACO08MAR]麻烦的干草打包机The Loathesome Hay Baler
P2903 [USACO08MAR]麻烦的干草打包机The Loathesome Hay Baler
题目描述
Farmer John has purchased the world's most loathesome hay baler. Instead of having a drive-roller that drives maybe an idler roller that drives the power take-off for the baler, it has N rollers (2 <= N <= 1050) which drive and are driven by various rollers.
FJ has meticulously cataloged data for each roller i: X_i,Y_i are the center of the roller (-5000 <= X_i <= 5000; -5000 <= Y_i <= 5000); R_i is the roller's radius (3 <= R_i <= 800). The drive-roller is located at 0,0; the baler power take-off is located at X_t,Y_t (numbers supplied in the input).
The drive-roller turns clockwise at 10,000 revolutions per hour. Your job is to determine the speeds of all the rollers that are in the power-train: from the drive-roller through the power take-off roller. Rollers that do not transfer power to the take-off roller are to be ignored. A roller of radius Rd that is turning at S rph and driving another roller of radius Rx will cause the second roller to turn at the speed -S*Rd/Rx (where the sign denotes whether the roller is turning clockwise or counterclockwise (anticlockwise for our British friends)).
Determine the power-train path and report the sum of the absolute values of all those rollers' speeds. All the rollers in the input set except the driver-roller are driven by some other roller; power is never transferred to a roller from more than one other roller.
Report your answer as an integer that is the truncated value after summing all the speeds.
Farmer John新买的干草打包机的内部结构大概算世界上最混乱的了，它不象普通的机器一样有明确的内部传动装置，而是，N (2 <= N <= 1050)个齿轮互相作用，每个齿轮都可能驱动着多个齿轮。 FJ记录了对于每个齿轮i，记录了它的3个参数：X_i，Y_i表示齿轮中心的位置坐标(-5000 <= X_i <= 5000; -5000 <= Y_i <= 5000)；R_i表示该齿轮的半径(3 <= R_i <= 800)。
驱动齿轮的位置为0,0，并且FJ也知道最终的工作齿轮位于X_t,Y_t。 驱动齿轮顺时针转动，转速为10,000转/小时。你的任务是，确定传动序列中所有齿轮的转速。传动序列的定义为，能量由驱动齿轮传送到工作齿轮的过程中用到的所有齿轮的集合。对能量传送无意义的齿轮都应当被忽略。
在一个半径为Rd，转速为S转/每小时的齿轮的带动下，与它相接的半径为Rx的齿轮的转速将为-S*Rd/Rx转/小时。S前的负号的意思是，一个齿轮带动的另一个齿轮的转向会与它的转向相反。
FJ只对整个传动序列中所有齿轮速度的绝对值之和感兴趣，你的任务也就相应转化成求这个值。机器中除了驱动齿轮以外的所有齿轮都被另外某个齿轮带动，并且不会出现2个不同的齿轮带动同一个齿轮的情况。
输入输出格式
输入格式：-
Line 1: Three space-separated integers: N, X_t, and Y_t
- Lines 2..N+1: Line i+1 describes roller i's properties: X_i, Y_i, and R_i
- Line 1: A single integer that is the truncated version of the sum of the absolute value of the speeds of the rollers in the power-train including the drive-roller, all the driven rollers, and the power take-off roller.
输入输出样例
4 32 54 0 0 10 0 30 20 32 54 20 -40 30 20
20000
说明
Four rollers: the drive-roller at 0,0 with radius 10. It drives the roller above it at 0,30 with radius 20. That roller drives both the power take-off roller at 32,54 (r=20) and a random roller (not in the power train) at -40,30 (r=20).
Roller Radius Speed
1 (0,0) 10 10,000
2 (0,30) 20 -5,000
3 (32,54) 20 5,000
Sum of abs values: 20,000
bfs一下就可以了。
注意几个问题：
1、相切才能联动！
2、精度！用double！
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <algorithm>
#include <queue>
#include <cmath>
const int sx = 1;
const int sy = 1;
const int K = 0;
const int MAXN = 1050 + 10;
int n,ex,ey;
int pre[MAXN];
int start;
int b[MAXN];
struct T
{
int x,y,r;
double energy, s;
}circle[MAXN];
inline bool IsPre(T& a, T& b)
{
int d = (a.x - b.x)*(a.x - b.x)+ (a.y - b.y)*(a.y - b.y);
int sum = a.r + b.r;
return d == sum*sum;
}
void bfs()
{
std::queue<T> q;
q.push(circle[start]);
b[start] = true;
while(!q.empty())
{
T temp = q.front();
q.pop();
for(int i = 1;i <= n;i ++)
{
if( !b[i] && IsPre(circle[i], temp))
{
b[i] = true;
circle[i].s = temp.s * temp.r / circle[i].r;
circle[i].energy = circle[i].s + temp.energy;
if(circle[i].x == ex && circle[i].y == ey)
{
printf("%d", (int)circle[i].energy);
return ;
}
q.push(circle[i]);
}
}
}
}
int main()
{
scanf("%d%d%d", &n ,&ex, &ey);
ex += K;
ey += K;
for(int i = 1;i <= n;i ++)
{
scanf("%d%d%d", &circle[i].x, &circle[i].y, &circle[i].r);
circle[i].x += K;
circle[i].y += K;
if(circle[i].x == K && circle[i].y == K)
{
start = i;
circle[i].s = 10000;
circle[i].energy = 10000;
}
}
bfs();
return 0;
}
