2732: [HNOI2012]射箭( 半平面交 )

很久没写题解了= =，来水一发吧= =

首先这道题很明显就是求y=ax^2+bx的是否有值取，每一个式子都代表着两个半平面，然后直接半平面交就行了

借鉴了hzwer的代码，还是特别简洁的说

CODE：

#include<cstdio>

#include<iostream>

#include<algorithm>

#include<cstring>

#include<cmath>

using namespace std;

typedef pair<long double,long double> ii;

typedef pair<ii,ii> line;

#define fi first

#define se second

#define maxn 101000

#define inf 1e15

#define exp 0

inline long double cross(ii x,ii y,ii z) {

return (x.fi-y.fi)*(x.se-z.se)-(x.se-y.se)*(x.fi-z.fi);

}

bool cmp(line x,line y) {return cross(x.fi,x.se,y.fi)>exp;}

line q[maxn*2];

long double tag[maxn*2];

bool cmp1(int x,int y) {if (tag[x]==tag[y]) return cmp(q[x],q[y]);return tag[x]<tag[y];} 

int l;

inline void init(){

q[1]=line(ii(-inf,-inf),ii(inf,-inf));

q[2]=line(ii(inf,-inf),ii(inf,inf));

q[3]=line(ii(inf,inf),ii(-inf,inf));

q[4]=line(ii(-inf,inf),ii(-inf,-inf));

l=4;

}

ii inter(line a,line b) {

long double k1,k2,t;

k1=cross(a.fi,b.se,a.se);

k2=cross(a.fi,a.se,b.fi);

t=k2/(k1+k2);

return ii(b.fi.fi+t*(b.se.fi-b.fi.fi),b.fi.se+t*(b.se.se-b.fi.se));

}

bool jud(line a,line b,line t) {

ii p=inter(a,b);

return cross(t.fi,p,t.se)>exp;

}

line a[maxn*2],deq[maxn*2];

int id[maxn*2];

inline bool check(int n) {

int num=0;

for (int i=1;i<=l;i++) if (id[i]<=n*2) a[++num]=q[id[i]];

int l,r;

deq[r=l=1]=a[1];

for (int i=2;i<=n*2;i++) {

while (r>l&&jud(deq[r-1],deq[r],a[i])) r--;

while (l<r&&jud(deq[l+1],deq[l],a[i])) l++;

deq[++r]=a[i];

}

while (l<r&&jud(deq[r-1],deq[r],deq[l])) r--;

while (l<r&&jud(deq[l+1],deq[l],deq[r])) l++;

return r-l>=2;

}

long double cal(long double a,long double b,long double x) {return b/a-a*x;}

int main(){

int n;

scanf("%d",&n);

init();

for (int i=1;i<=n;i++) {

int x1,y1,y2;

scanf("%d%d%d",&x1,&y1,&y2);

q[++l]=line(ii(-1,cal(x1,y1,-1)),ii(1,cal(x1,y1,1)));

q[++l]=line(ii(1,cal(x1,y2,1)),ii(-1,cal(x1,y2,-1)));

}

for (int i=1;i<=n*2+4;i++) {tag[i]=atan2(q[i].se.se-q[i].fi.se,q[i].se.fi-q[i].fi.fi);id[i]=i;}

sort(id+1,id+1+n*2+4,cmp1); 

int l=2,r=n+2;

while (l+1<r) {

int mid=(l+r)>>1;

if (check(mid)) l=mid;

else r=mid;

}

printf("%d\n",check(r)?r-2:l-2);

return 0;

}