二分 Codeforces591D Chip 'n Dale Rescue Rangers

传送门：点击打开链接

题意：前t秒，风速的向量为(vx,vy)，t秒后风速的向量变成(wx,wy)，刚开始在(x1,y1)，要去(x2,y2)，飞机速度的最大大小是v，问到达目的地至少需要多长的时间

思路：二分时间，那么就能求出风速对飞机位移的影响，那么影响后，现在与目的地的距离求出来，让飞机直行看是否能在时间内到达，就算验证二分是否满足条件了，剩下的二分就行。

#include<map>
#include<set>
#include<cmath>
#include<ctime>
#include<stack>
#include<queue>
#include<cstdio>
#include<cctype>
#include<string>
#include<vector>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<functional>
#define fuck(x) cout<<"["<<x<<"]"
#define FIN freopen("input.txt","r",stdin)
#define FOUT freopen("output.txt","w+",stdout)
using namespace std;
typedef long long LL;

int main() {
    double x1, y1, x2, y2; //FIN;
    double v, t, vx, vy, wx, wy;
    while(~scanf("%lf%lf%lf%lf%lf%lf%lf%lf%lf%lf", &x1, &y1, &x2, &y2, &v, &t, &vx, &vy, &wx, &wy)) {
        double l = 0, r = 1e9, m;
        for(int i = 1; i <= 100; i++) {
            m = (l + r) / 2;
            double x = x1 + vx * min(m, t) + wx * max(m - t, 0.0);
            double y = y1 + vy * min(m, t) + wy * max(m - t, 0.0);
            double dist = sqrt((x - x2) * (x - x2) + (y - y2) * (y - y2));
            if(dist <= v * m) r = m;
            else l = m;
        }
        printf("%.10lf\n", l);
    }
    return 0;
}