BNU 34986 Football on Table

"Bored?

Let‘s play table football!"

The table football is played on a rectangular table, usually contains m rows of players which are plastic, metal, wooden, or sometimes carbon-fibre figures mounted on vertical metal bars. After
playing table football for hours, we decide to take a rest. And the state of the table remains random, that means each bar is placed at any legal position with equal possibilities (players can’t be outside the table and a bar is fixed at a row).

Now I‘m wondering if the goal-keeper shoot a ball, what’s the possibility of this shoot turning to a goal?

(If the ball did not touch any player, then I made a goal).

Let‘s assume there is a_i players on the i^th row
(counted from left to right). And we know the width of each player and the distance between two players. (To simplify the problem, we ignore the thickness of the players, in other words, we consider the players as vertical segments. Then we treat the football
as a point, moving along a straight line and will not touch the boundary of the table).

Input

The first line contains an integer T, which denotes the number of test cases.

For each test case:

• The first line contains two numbers L, W (1 ≤ L, W ≤ 10⁸), denoting the length and the width of the table. (the lower left corner of the table is (0, 0) , and the top right corner of the
  table is (L, W)).
• The second line contains four number X, Y, d_x, d_y. (X, Y) denotes the initial position of the ball and (d_x,
  d_y) denotes the shooting direction. (X will always be zero, 0 ≤ Y ≤ W, d_x> 0).
• The third line contains an integer m (1 ≤ m ≤ 10), the number of rows of the players.
• Following m blocks, for the i^th block,
  • The first line contains a number x_i and an integer a_i,(0<x_i<L,
    1 ≤ a_i ≤ 100) denoteing the x-coordinate of the i^th row and the number of players at the i^th row.
  • The second line contains a_i numbers, the j^th number w_j denotes
    the width of the j^th (from bottom to top) player at the i^th row.
  • The third line contains a_i - 1 numbers, the j^th number d_j denotes
    the distance between the j^th player and the (j+1)^th player. If a_i equals
    1, this line will be a blank line.

We guarantee that ∑w_j + ∑d_j + 1< W

Output

For each case, first output the case number as "Case #x: ", and x is the case number. Then output the result rounded to 5
digits after the decimal point, representing the possibility of this shoot turning to a goal, in other words, that the ball does not touch any player.

Sample Input

2
8.0 10.0
0.0 5.0 2.0 -0.1
1
3.0 2
2.0 2.0
1.0
8.0 10.0
0.0 5.0 2.0 0.0
2
3.0 2
2.0 2.0
1.0
4.0 3
2.0 1.0 2.0
1.0 1.0

Sample Output

Case #1: 0.23000
Case #2: 0.13333

Hint

The black solid lines denote the table.

The dashed line denotes the bar.

The gray lines denote the players.

The dot-dashed line denote the trajectory of the ball.

Source

2014
ACM-ICPC Beijing Invitational Programming Contest

题意：：一个人在玩桌面足球，有m行球员。每行球员有ai个，给出每一个球员的宽度和相邻球员之间的距离，球从最左边射出，给出球的起点坐标跟方向向量，问可以到达最右边的概率。

思路：看懂题意就好做点了，枚举每行能够碰到球的距离。然后概率求反

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
const int maxn = 105;

double w[maxn], tail[maxn], pos[maxn];

int main() {
	int t, m, cas = 1;
	double W, L, X, Y, dx, dy, dis;
	scanf("%d", &t);
	while (t--) {
		double ans = 1.0;
		scanf("%lf%lf", &L, &W);
		scanf("%lf%lf%lf%lf", &X, &Y, &dx, &dy);
		scanf("%d", &m);

		while (m--) {
			double sum = 0.0;
			pos[0] = 0.0;
			double x;
			int n;
			scanf("%lf%d", &x, &n);
			double y = Y + dy * (x - X) / dx;
			for (int i = 0; i < n; i++) {
				scanf("%lf", &w[i]);
				sum += w[i];
			}
			tail[0] = w[0];
			for (int i = 1; i < n; i++) {
				scanf("%lf", &dis);
				pos[i] = pos[i-1] + w[i-1] + dis;
				tail[i] = pos[i] + w[i];
				sum += dis;
			}
			double cnt = 0.0, len = 0.0;
			double mv = W - sum;
			for (int i = 0; i < n; i++) {
				cnt = 0.0;
				if ((pos[i] <= y) && (tail[i] + mv) >= y) {
					if (tail[i] >= y)
						cnt = (pos[i] + mv <= y) ? mv : (y - pos[i]);
					else cnt = (pos[i] + mv >= y) ?

w[i] : (tail[i] - y + mv);
				}
				len += cnt;
			}
			if (mv == 0.0) {
				ans = 0;
				break;
			}
			else ans = ans * (mv - len) / mv;
		}

		printf("Case #%d: %.5lf\n", cas++, ans);
	}
	return 0;
}