SDOI2018 物理實驗

SDOI2018 物理實驗

題意：二維平面上有一條直線，直線上放置了一個激光發射器，會嚮導軌兩側沿導軌垂直方向發射寬度爲 L 的激光束。平面上還有 n 條線段，而且線段和線段、線段和直線之間都沒有公共點，線段和直線的夾角不超過 85◦，激光束不能穿透線段，你須要求出激光束能照射到的線段長度之和的最大值。
作法：先進行座標變換，將給定直線移到x軸上，注意首先將直線的一端移到原點，而後再旋轉。（樣例數據全都從原點出發。。。沒發現。就gg了十幾發，好腦殘啊。。）考慮將每一個線段投影到直線上，這些點將直線分紅一些線段，對於每一段，找到它上方最近的那個線段，而後計算單位長度的貢獻，最後的答案就能夠雙指針求出了。對於查詢上方最近的線段，使用set維護掃描線便可。雙指針還寫崩了。。。這題出出來就gg了啊。。。碼力弱炸，還手殘。

#include <bits/stdc++.h>
using namespace std;
#define rep(i,a,b) for(int i=(a);i<=(b);++i)
#define per(i,a,b) for(int i=(a);i>=(b);--i)
#define pb push_back

typedef long double db;

const db EPS = 0;
const db INF = 1e100;
inline int sgn(db x, db E = EPS){return x<-E?-1:x>E;}
inline int cmp(db a,db b,db E = EPS){return sgn(a-b, E);}

struct P{
    db x,y;
    bool operator<(P o) const { return cmp(x,o.x) != 0? cmp(x,o.x) == -1 : cmp(y,o.y) == -1; }
    bool operator>(P o) const { return cmp(x,o.x) != 0? cmp(x,o.x) == 1 : cmp(y,o.y) == 1; }
    bool operator==(P o) { return cmp(x,o.x) == 0 && cmp(y,o.y) == 0; }
    void read(){ int _x,_y; scanf("%d%d",&_x,&_y); x=_x,y=_y; }
    P operator + (P o) { return {x+o.x,y+o.y}; }
    P operator - (P o) { return {x-o.x,y-o.y}; }
    db det(P o) { return x*o.y-y*o.x; }
    db dot(P o){ return x*o.x+y*o.y; }
    db disTo(P p) { return (*this - p).abs(); }
    db alpha() { return atan2(y, x); }
    db abs() { return sqrt(abs2());}
    db abs2() { return x * x + y * y; }
    P rot(db a) {
        db c = cos(a), s = sin(a);
        return {c*x-s*y, s*x+c*y};
    }
};

struct L {
    P ps[2];
    L(){}
    L(P p1, P p2) { ps[0] = p1, ps[1] = p2; }
    P& operator[](int i) { return ps[i]; }
    P dir() { return ps[1] - ps[0]; }
    bool include(P p) { return sgn((ps[1]-ps[0]).det(p-ps[0])); }
};

db nowx = 0;
int top[100010];
vector<db> scan;
int sc;

struct seg {
    P u, v; int id; db k;
    seg(){}
    seg(P _u, P _v, int _id) {
        id = _id; u = _u; v = _v;
        k = (v.y - u.y) / (v.x - u.x);
    }
    bool operator < (const seg a) const {
        db v1 = u.y + k * (nowx - u.x);
        db v2 = a.u.y + a.k * (nowx - a.u.x);
        if( cmp(v1, v2) == 0 ) return k < a.k;
        return v1 < v2;
    }
};
set< seg > S;
set< seg >::iterator it[100010];

struct Q {
    P e; int id;
    bool operator < (Q a) {
        if(e == a.e) return id < a.id;
        return e < a.e;
    }
};
vector< Q > qs;
vector<db> cal(vector<L> ls) {
    int n = ls.size();
    qs.clear();
    rep(i,0,n-1) {
        if(ls[i][0].x > ls[i][1].x) swap(ls[i][0], ls[i][1]);
        qs.pb({ls[i][0], -i-1});
        qs.pb({ls[i][1], i+1});
    }
    sort(qs.begin(),qs.end());
    nowx = 0; S.clear();
    rep(i,0,sc-1) top[i] = -1;
    for(int j = 0, i = 0; i < sc; ++ i) {
        nowx = scan[i];
        for( ; j < qs.size() && qs[j].e.x <= nowx ; ++j) {
            if(qs[j].id < 0) {
                int ID = (-qs[j].id)-1;
                it[ID] = S.insert(seg(ls[ID][0],ls[ID][1],ID)).first;
            }
            else {
                int ID = qs[j].id-1;
                S.erase(it[ID]);
            }
        }
        if(!S.empty()) top[i] = (*S.begin()).id;
    }
    function<db(int)> cost = [&](int tp) -> db {
        return abs( ls[tp][0].disTo(ls[tp][1]) / (ls[tp][1].x - ls[tp][0].x) );
    };
    vector<db> V;
    rep(i, 0, sc-2) {
        if(top[i] != -1) V.pb(cost(top[i])); else V.pb(0);
    }
    return V;
}

db pv[100010], pc[100010];

db cala(vector< pair<db,db> > &V, db Len) {
    function<db(db,int,int)> MX = [&](db LL, int p1, int p2) -> db {
        db tmp = 0;
        if(0 <= p1 && p1 <= sc-2) tmp = max(tmp, V[p1].second * min(LL, V[p1].first) );
        if(0 <= p2 && p2 <= sc-2) tmp = max(tmp, V[p2].second * min(LL, V[p2].first) );
        return tmp;
    };
    db ans = 0;
    int p1 = 0, p2 = 0; db tC = 0, tL = 0;
    do {
        if(Len >= tL) ans = max(ans, tC + MX(Len-tL, p1-1, p2 ) );
        if(p2 < sc-1 && cmp(tL + V[p2].first, Len) <= 0) {
            tL += V[p2].first;
            tC += V[p2].first * V[p2].second;
            ++ p2;
        }
        else if(p1 <= p2) {
            tL -= V[p1].first;
            tC -= V[p1].first * V[p1].second;
            ++ p1;
        }
        else {
            if(p2 < sc-1) {
                tL += V[p2].first;
                tC += V[p2].first * V[p2].second;
                ++ p2;
            }
            if(p1 <= p2) {
                tL -= V[p1].first;
                tC -= V[p1].first * V[p1].second;
                ++ p1;
            }
        }
        if(Len >= tL) ans = max(ans, tC + MX(Len-tL, p1-1, p2 ) );
    } while(p1 < sc-1 || p2 < sc-1);
    return ans;
}

int n;
L line[100010];
P p1, p2;
db Len;
vector<L> UP, LW;

void init() {
    UP.clear(); LW.clear();
    if(p1 > p2) swap(p1, p2);
    db a = -(p2-p1).alpha();
    rep(i,0,n-1) {
        line[i][0] = (line[i][0]-p1).rot(a);
        line[i][1] = (line[i][1]-p1).rot(a);
        if( sgn(line[i][0].y) == 1 ) UP.pb(line[i]);
        else LW.pb( L( {line[i][0].x, -line[i][0].y} , {line[i][1].x, -line[i][1].y} ) );
    }
    scan.clear();
    rep(i,0,(int)UP.size()-1) { scan.pb(UP[i][0].x), scan.pb(UP[i][1].x); }
    rep(i,0,(int)LW.size()-1) { scan.pb(LW[i][0].x), scan.pb(LW[i][1].x); }
    sort(scan.begin(),scan.end()); scan.erase( unique(scan.begin(),scan.end()), scan.end());
    sc = scan.size();
}

int main() {
    int _; scanf("%d",&_);
    while(_--) {
        scanf("%d",&n);
        rep(i,0,n-1) {
            line[i][0].read();
            line[i][1].read();
        }
        p1.read(), p2.read();
        int t; scanf("%d",&t), Len = t;

        init();

        vector<db> V1 = cal(UP);
        vector<db> V2 = cal(LW);

        vector<pair<db,db>> V;
        rep(i, 0, (int)V1.size()-1) {
            V.pb({scan[i+1]-scan[i], V1[i] + V2[i]});
        }
        db ans = cala(V, Len);
        printf("%.12Lf\n",ans);
    }
    return 0;
}