解析
暴力求出所有三角形之間的所有交點，提出所有的橫坐標。
然后任意兩個相鄰的橫坐標之間的面積都是若干個梯形。
那么就可以求出對于每一個橫坐標截得的三角形長度的并的和，然后加在一起乘高除以二即可。
在這里插入代碼片`#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define ull unsigned long long
#define debug(...) fprintf(stderr,__VA_ARGS__)
//#define ok debug("OK\n")
inline ll read() {ll x(0),f(1);char c=getchar();while(!isdigit(c)) {if(c=='-') f=-1;c=getchar();}while(isdigit(c)) {x=(x<<1)+(x<<3)+c-'0';c=getchar();}return x*f;
}
const int N=2e5+100;
const double eps=1e-10;
const double inf=1e9;int n,m;struct pt {double x,y;
} p[N];
inline pt operator + (const pt a,const pt b) {return (pt) {a.x+b.x,a.y+b.y};
}
inline pt operator - (const pt a,const pt b) {return (pt) {a.x-b.x,a.y-b.y};
}
inline pt operator * (const double k,const pt a) {return (pt) {k*a.x,k*a.y};
}
inline pt operator / (const pt a,const double k) {return (pt) {a.x/k,a.y/k};
}
inline pt operator * (const pt a,const double k) {return (pt) {k*a.x,k*a.y};
}
inline double operator * (const pt a,const pt b) {return a.x*b.x+a.y*b.y;
}
inline double operator ^ (const pt a,const pt b) {return a.x*b.y-b.x*a.y;
}
inline double len (const pt a) {return sqrt(a.x*a.x+a.y*a.y);
}
inline pt danwei(const pt a) {return a/len(a);
}
inline pt chui(const pt a) {return (pt) {a.y,-a.x};
}
void print(pt o,int op=1) {printf("%.2lf %.2lf",o.x,o.y);putchar(op?'\n':' ');return;
}
void input(pt &o) {scanf("%lf%lf",&o.x,&o.y);
}struct line {pt a,b,d;
};void input(line &l) {double a,b,c,d;scanf("%lf%lf%lf%lf",&a,&b,&c,&d);l=(line) {(pt) {a,b},(pt) {c,d},(pt) {c-a,d-b}};return;
}
void print(line l,int op=1){print(l.a,0);print(l.b,op);
}
inline line trans(double a,double b,double c,double d) {return (line) {(pt) {a,b},(pt) {c,d},(pt) {c-a,d-b}};
}
inline line trans(const pt a,const pt b) {return (line) {a,b,b-a};
}
int pos(const line l1,const line l2) {if(abs(l1.d^l2.d)>eps) return 1;else if(abs((l2.a-l1.a)^l1.d)>eps) return 2;else return 3;
}
pt jiaodian(const line l1,const line l2) {double k=((l2.a-l1.a)^l2.d)/(l1.d^l2.d);return l1.a+(k*l1.d);
}
pt a[N],b[N],c[N];
line l[N][4];
double q[N];
int cnt;
bool ok(line l,pt o){return (l.a.x-eps<o.x&&o.x<l.b.x+eps)||(l.b.x-eps<o.x&&o.x<l.a.x+eps);
}
struct seg{double l,r;bool operator < (const seg oth)const{return l<oth.l;}
}s[N];
int num;
line L;
double w[5];
void work(int k){int jd(0);//printf("work : %d\n",k);for(int i=1;i<=3;i++){if(abs(l[k][i].d.x)<eps) continue;pt o=jiaodian(l[k][i],L);//print(o,1);if(ok(l[k][i],o)) w[++jd]=o.y;}if(jd>1){++num;s[num].l=s[num].r=w[1];for(int i=2;i<=jd;i++) s[num].l=min(s[num].l,w[i]);for(int i=2;i<=jd;i++) s[num].r=max(s[num].r,w[i]);//printf("(%.2lf %.2lf)\n",s[num].l,s[num].r);}
}
double calc(double x){num=0;L=trans(x,0,x,1);for(int i=1;i<=n;i++) work(i);sort(s+1,s+1+num);double l=-inf,r=-inf,res=0;for(int i=1;i<=num;i++){if(s[i].l<r+eps) r=max(r,s[i].r);else{res+=r-l;l=s[i].l;r=s[i].r;}}res+=r-l;return res;
}
signed main() {
#ifndef ONLINE_JUDGE//freopen("a.in","r",stdin);//freopen("a.out","w",stdout);
#endifreturn 0;
}
/*
2
0.0 0.0 20.0 28.0 10.0 35.0
-50.0 0.0 -70.8 -33.5 -30.5 -10.5
*/`
總結
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