SDOI2018 物理實(shí)驗(yàn) 題意：二維平面上有一條直線，直線上放置了一個(gè)激光發(fā)射器，會(huì)向?qū)к墐蓚?cè)沿導(dǎo)軌垂直方向發(fā)射寬度為 L 的激光束。平面上還有 n 條線段，并且線段和線段、線段和直線之間都沒有公共點(diǎn)，線段和直線的夾角不超過 85?，激光束不能穿透線段，你需要求出激光束能照射到的線段長度之和的最大值。一端移到原點(diǎn) ，然后再旋轉(zhuǎn)。（樣例數(shù)據(jù)全都從原點(diǎn)出發(fā)。。。沒發(fā)現(xiàn)。就gg了十幾發(fā)，好腦殘啊。。）考慮將每個(gè)線段投影到直線上，這些點(diǎn)將直線分成一些線段，對(duì)于每一段，找到它上方最近的那個(gè)線段，然后計(jì)算單位長度的貢獻(xiàn)，最后的答案就可以雙指針求出了。對(duì)于查詢上方最近的線段，使用set維護(hù)掃描線即可。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_backtypedef 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;
}
轉(zhuǎn)載于:https://www.cnblogs.com/RRRR-wys/p/10952057.html
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