bnu 34986 Football on Table(數學+暴力)

題目連接：bnu 34986 Football on Table

題目大意:給出桌子的大小L，W，然後是球的起始位置sx，sy，以及移動的向量dx，dy，然後給出n，表示有n個桿，對於每個桿，先給出位置x，以及桿上有多少個小人c，給出小人的寬度，再給出c個小人間的距離。現在問說球有多少個概率可以串過所有人。

解題思路；對於每個桿求無阻擋的概率，注意概率 = 空隙 / 可移動的范圍大小，而不是W。其他就水水的。

#include 
#include 
#include 
#include 

using namespace std;
const double eps = 1e-8;
const int N = 205;

double L, W;
double sx, sy, dx, dy;
double d[N];

double solve () {

    int n, c;
    double w, ans = 1;

    scanf("%d", &n);

    while (n--) {
        scanf("%lf%d", &w, &c);
        double r = sy + w * dy / dx;
        double s = 0;
        c = 2 * c - 1;

        for (int i = 0; i < c; i += 2)
            scanf("%lf", &d[i]);
        for (int i = 1; i < c; i += 2)
            scanf("%lf", &d[i]);    
        for (int i = 0; i < c; i++)
            s += d[i];

        double l = W - r;
        double rec = 0;

        if (l > s) {
            rec += (l - s);
            l = 0;
        } else {
            l = s - l;
        }

        if (r > s) {
            rec += (r - s);
            r = s;
        }


        s = 0;
        for (int i = 0; i < c; i++) {
            double tmp = s + d[i];
            if (i&1) {
                double add = min(r, tmp) - max(s, l);
                rec += max(add, (double)0);
            }
            s = tmp;
        }

        ans *= rec / (W-s);
    }
    return ans;
}

int main () {
    int cas;
    scanf("%d", &cas);
    for (int i = 1; i <= cas; i++) {
        scanf("%lf%lf", &L, &W);
        scanf("%lf%lf%lf%lf", &sx, &sy, &dx, &dy);
        printf("Case #%d: %.5lf\n", i, solve());
    }
    return 0;
}