pid=34986" target="_blank" style="">题目连接：bnu 34986 Football on Table

题目大意:给出桌子的大小L，W，然后是球的起始位置sx，sy，以及移动的向量dx，dy。然后给出n。表示有n个杆，对于每一个杆。先给出位置x，以及杆上有多少个小人c，给出小人的宽度。再给出c个小人间的距离。如今问说球有多少个概率能够串过全部人。

解题思路。对于每一个杆求无阻挡的概率。注意概率 = 空隙 / 可移动的范围大小，而不是W。其它就水水的。

#include <cstdio>

#include <cstring>

#include <cmath>

#include <algorithm>

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;

}