SDOI2018物理实验

/*
向量运算不会呐
抄了一个长度几百行的模板  一直过不了编译  醉了
还是抄了大佬的代码

首先把所有的线段投影到 导轨上  然后用set 分上和下分别维护一下 距离导轨最近的线段  是能够照射到的

可以证明 我们的最优答案有一端肯定是在线段的分界点上的  所以我们可以用扫描线思想 
从一端扫到另一端  端点为各个分界点  这样正这反着处理两遍即可 


long double  怎么输出啊   在线等挺急的 
*/




#include<cstdio>
#include<algorithm>
#include<cstring>
#include<iostream>
#include<cmath>
#include<set>
#define sqr(x) (x) * (x)
#define ld long double
#define ll long long
#define M 20010
using namespace std;

int read() {
    int nm = 0, f = 1;
    char c = getchar();
    for(; !isdigit(c); c = getchar()) if(c == '-') f = -1;
    for(; isdigit(c); c  =getchar() ) nm = nm * 10 + c - '0';
    return nm * f;
}
const ld eps = 1e-9, zz = 0;
ld x[M][2], y[M][2], len[M] ,ver[M], z[M], x0;
int p[M] , n,m;

bool cmp(int a, int b) {
    return (a > 0 ? x[a][0] : x[-a][1]) < (b > 0 ? x[b][0]: x[-b][1]);
}

struct L {
    int t;
    bool operator <( const L &b) const {
        ld w1 = (y[t][1] - y[t][0]) / (x[t][1] - x[t][0]) * (x0 - x[t][0])+y[t][0];
        ld w2 = (y[b.t][1] - y[b.t][0]) / (x[b.t][1] - x[b.t][0]) * (x0 - x[b.t][0]) + y[b.t][0];
        return fabs(w1) < fabs(w2);
    }
};
set<L>up, down;
int main() {
    int t = read();
    while(t--) {
        n = read();
        for(int i = 1; i <= n; i++) {
            x[i][0] = read(), y[i][0] = read(), x[i][1] = read(), y[i][1] = read();
            len[i] = sqrt(sqr(x[i][0] - x[i][1]) + sqr(y[i][0] - y[i][1]));
        }
        ld xn0 = read(), yn0 = read(), xn1 = read(), yn1 = read(), lenx = read();
        if(xn0 > xn1) swap(xn0, xn1), swap(yn0, yn1);
        ld dx = xn1 - xn0, dy = yn1 - yn0, le = sqrt(sqr(dx) + sqr(dy)),Sin = dy / le, Cos = dx / le;
        for(int i = 1; i <= n; i++) {
            x[i][0] -= xn0, x[i][1] -= xn0, y[i][1] -= yn0, y[i][0] -= yn0;
            ld a,b,c,d;
            a = x[i][0] * Cos + y[i][0] * Sin;
            b = x[i][0] * Sin - y[i][0] * Cos;
            c = x[i][1] * Cos + y[i][1] * Sin;
            d = x[i][1] * Sin - y[i][1] * Cos;
            x[i][0] = a, y[i][0] = -b, x[i][1] = c, y[i][1] = -d;
        }
        for(int i = 1; i <= n; i++) {
            if(x[i][0] > x[i][1]) swap(x[i][0], x[i][1]), swap(y[i][0], y[i][1]);
            len[i] = len[i] /(x[i][1] - x[i][0]);
        }
        m = 0;
        for(int i = 1; i <= n; i++) p[++m] = i, p[++m] = -i;
        sort(p + 1, p + m + 1, cmp);
        for(int i = 1; i <= m ; i++) ver[i] = 0;
        for(int i = 1, a; i <= m; i++) {
            if(p[i] > 0) {
                a = p[i], z[i] = x0 = x[a][0];
                if(y[a][0] > 0) up.insert((L){a});
                else down.insert((L){a});
            }
            else{
                a = -p[i], z[i] = x0 = x[a][1];
                if(y[a][0] > 0) up.erase((L){a});
                else down.erase((L){a});
            }
            if(!up.empty()) ver[i] += len[(*up.begin()).t];
            if(!down.empty()) ver[i] += len[(*down.begin()).t];
        }
        ld ans = 0, sum = 0; z[m + 1] = 1e16;
        for(int i = 1, l = 1; i <= m; i++){
            sum += (z[i] - z[i - 1]) * ver[i - 1];
            while(z[i] - z[l + 1] > lenx) l++, sum -= (z[l] - z[l - 1]) * ver[l - 1];
            ans = max(ans, sum - max(zz, (z[i] - z[l] - lenx) * ver[l]));
        }
        sum = 0;
        for(int i = m, r = m; i >= 1; i--)
        {
            sum += (z[i + 1] - z[i]) * ver[i];
            while(z[r - 1] - z[i] > lenx) --r, sum -= (z[r + 1] - z[r]) * ver[r];
            ans = max(ans, sum - max(zz, (z[r] - z[i] - lenx) * ver[r - 1]));
        }
        //cout << ans << "\n";
        double as = ans;
        printf("%.15lf\n", as);
    }
    return 0;
    }