bzoj 1857: [Scoi2010]传送带 三分

题目链接

1857: [Scoi2010]传送带

Time Limit: 1 Sec  Memory Limit: 64 MB
Submit: 934  Solved: 501
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Description

在一个2维平面上有两条传送带，每一条传送带可以看成是一条线段。两条传送带分别为线段AB和线段CD。lxhgww在AB上的移动速度为P，在CD上的移动速度为Q，在平面上的移动速度R。现在lxhgww想从A点走到D点，他想知道最少需要走多长时间

Input

输入数据第一行是4个整数，表示A和B的坐标，分别为Ax，Ay，Bx，By 第二行是4个整数，表示C和D的坐标，分别为Cx，Cy，Dx，Dy 第三行是3个整数，分别是P，Q，R

Output

输出数据为一行，表示lxhgww从A点走到D点的最短时间，保留到小数点后2位

Sample Input

0 0 0 100
100 0 100 100
2 2 1

Sample Output

136.60

 

三分套三分就可以了

#include <iostream>
#include <vector>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <map>
#include <set>
#include <string>
#include <queue>
#include <stack>
#include <bitset>
using namespace std;
#define pb(x) push_back(x)
#define ll long long
#define mk(x, y) make_pair(x, y)
#define lson l, m, rt<<1
#define mem(a) memset(a, 0, sizeof(a))
#define rson m+1, r, rt<<1|1
#define mem1(a) memset(a, -1, sizeof(a))
#define mem2(a) memset(a, 0x3f, sizeof(a))
#define rep(i, n, a) for(int i = a; i<n; i++)
#define fi first
#define se second
typedef pair<int, int> pll;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int mod = 1e9+7;
const int inf = 1061109567;
const int dir[][2] = { {-1, 0}, {1, 0}, {0, -1}, {0, 1} };
double xa, xb, xc, xd, ya, yb, yc, yd, p, q, r;
double dis(double x1, double y1, double x2, double y2) {
    return sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2));
}
double ternary(double x, double y) {
    double lx = xc, ly = yc, rx = xd, ry = yd;
    while(fabs(rx-lx)>eps || fabs(ry-ly)>eps) {
        double x1 = lx+(rx-lx)/3, x2 = lx+(rx-lx)/3*2;
        double y1 = ly+(ry-ly)/3, y2 = ly+(ry-ly)/3*2;
        double tmp1 = dis(x, y, x1, y1)/r+dis(x1, y1, xd, yd)/q+dis(x, y, xa, ya)/p;
        double tmp2 = dis(x, y, x2, y2)/r+dis(x2, y2, xd, yd)/q+dis(x, y, xa, ya)/p;
        if(tmp1>tmp2) {
            lx = x1, ly = y1;
        } else {
            rx = x2, ry = y2;
        }
    }
    return dis(x, y, lx, ly)/r+dis(lx, ly, xd, yd)/q+dis(x, y, xa, ya)/p;
}
double solve() {
    double lx = xa, rx = xb, ly = ya, ry = yb;
    while(fabs(rx-lx)>eps || fabs(ry-ly)>eps) {
        double x1 = lx+(rx-lx)/3, x2 = lx+(rx-lx)/3*2;
        double y1 = ly+(ry-ly)/3, y2 = ly+(ry-ly)/3*2;
        double tmp1 = ternary(x1, y1), tmp2 = ternary(x2, y2);
        if(tmp1>tmp2) {
            lx = x1, ly = y1;
        } else {
            rx = x2, ry = y2;
        }
    }
    return ternary(lx, ly);

}
int main()
{
    cin>>xa>>ya>>xb>>yb>>xc>>yc>>xd>>yd>>p>>q>>r;
    double ans = solve();
    printf("%.2f\n", ans);
    return 0;
}