Problem E: Slalom

You may select fromSpairs of skis, where thejthpair has speedsj. Your movement is governed by the following rule: if you select a pair of skis with speedsj, you move with a constant downward velocity ofsjmetres per second. Additionally, at any time you may move at a horizontal speed of at mostvhmetres per second.
You may start and finish at any two horizontal positions. Determine which pair of skis will allow you to get through the race course, passing through all the gates, in the shortest amount of time.
Input Specification
The first line of input contains a single integer, the number of test cases to follow.The first line of each test case contains the three integersW,vh, andN, separated by spaces, with 1 <=W<= 108, 1 <=vh<= 106, and 1 <=N<= 105.
The followingNlines of the test case each contain two integersxiandyi, the horizontal and vertical positions respectively of theithleft gate, with 1 <=xi,yi<= 108.
The next line of the test case contains an integerS, the number of skis, with 1 <=S<= 106.
The followingSlines of the test case each contain one integersj, the speed of thejthpair of skis, with 1 <=sj<= 106.
Sample Input
2 3 2 3 1 1 5 2 1 3 3 3 2 1 3 2 3 1 1 5 2 1 3 1 3
Output Specification
Output one line for each test case. If it is impossible to complete the race with any pair of skis, print the line IMPOSSIBLE . Otherwise, print the vertical speed sj of the pair of skis that allows you to get through the race course in the shortest time.Output for Sample Input
2 IMPOSSIBLE
#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
const int N = 100005;
int T, n, S, s[N * 10];
double w, v;
struct Sk {
double s, e, y;
} sk[N];
void init() {
scanf("%lf%lf%d", &w, &v, &n);
for (int i = 0; i < n; i ++) {
scanf("%lf%lf", &sk[i].s, &sk[i].y);
sk[i].e = sk[i].s + w;
}
scanf("%d", &S);
for (int i = 0; i < S; i ++)
scanf("%d", &s[i]);
sort(s, s + S);
}
bool judge(int sv) {
double s = sk[n - 1].s, e = sk[n - 1].e;
for (int i = n - 1; i > 0; i --) {
double time = (sk[i].y - sk[i - 1].y) / sv;
s = max(s - time * v, sk[i - 1].s);
e = min(e + time * v, sk[i - 1].e);
if (s > sk[i - 1].e || e < sk[i - 1].s || s > e)
return false;
}
return true;
}
void solve() {
if (!judge(s[0])) {
printf("IMPOSSIBLE\n");
return;
}
int l = 0, r = S - 1;
while (l < r) {
int mid = (l + r)>>1;
if (judge(s[mid])) l = mid + 1;
else r = mid;
}
if (!judge(s[l]))
l--;
printf("%d\n", s[l]);
}
int main() {
scanf("%d", &T);
while (T--) {
init();
solve();
}
return 0;
}