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传送门：csu 1812: 三角形和矩形
思路：首先，求出三角形的在矩形区域的顶点，矩形在三角形区域的顶点。然后求出所有的交点。这些点构成一个凸包，求凸包面积就OK了。
/************************************************************** Problem: User: youmi Language: C++ Result: Accepted Time: Memory: ****************************************************************/ //#pragma comment(linker, "/STACK:1024000000,1024000000") //#include<bits/stdc++.h> #include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <map> #include <stack> #include <set> #include <sstream> #include <cmath> #include <queue> #include <deque> #include <string> #include <vector> #define zeros(a) memset(a,0,sizeof(a)) #define ones(a) memset(a,-1,sizeof(a)) #define sc(a) scanf("%d",&a) #define sc2(a,b) scanf("%d%d",&a,&b) #define sc3(a,b,c) scanf("%d%d%d",&a,&b,&c) #define scs(a) scanf("%s",a) #define sclld(a) scanf("%I64d",&a) #define pt(a) printf("%d\n",a) #define ptlld(a) printf("%I64d\n",a) #define rep(i,from,to) for(int i=from;i<=to;i++) #define irep(i,to,from) for(int i=to;i>=from;i--) #define Max(a,b) ((a)>(b)?(a):(b)) #define Min(a,b) ((a)<(b)?(a):(b)) #define lson (step<<1) #define rson (lson+1) #define eps 1e-6 #define oo 0x3fffffff #define TEST cout<<"*************************"<<endl const double pi=4*atan(1.0); using namespace std; typedef long long ll; template <class T> inline void read(T &n) { char c; int flag = 1; for (c = getchar(); !(c >= ‘0‘ && c <= ‘9‘ || c == ‘-‘); c = getchar()); if (c == ‘-‘) flag = -1, n = 0; else n = c - ‘0‘; for (c = getchar(); c >= ‘0‘ && c <= ‘9‘; c = getchar()) n = n * 10 + c - ‘0‘; n *= flag; } ll Pow(ll base, ll n, ll mo) { if (n == 0) return 1; if (n == 1) return base % mo; ll tmp = Pow(base, n >> 1, mo); tmp = (ll)tmp * tmp % mo; if (n & 1) tmp = (ll)tmp * base % mo; return tmp; } //*************************** int n; const int maxn=100000+10; const ll mod=1000000007; double xx[10],yy[10]; int sgn(double x) { if(fabs(x)<eps) return 0; if(x<0) return -1; else return 1; } struct point { double x,y; point(){}; point(double _x,double _y) { x=_x,y=_y; } point operator-(const point &_b)const { return point(x-_b.x,y-_b.y); } double operator *(const point &_b)const { return x*_b.x+y*_b.y; } double operator^(const point &_b)const { return x*_b.y-_b.x*y; } bool operator==(const point &_b)const { return sgn(x-_b.x)==0&&sgn(y-_b.y)==0; } }; point tri[10],rec[10]; double dist(point a,point b) { return sqrt((a-b)*(a-b)); } struct line { point s, e; line() {} line(point _s, point _e) { s = _s; e = _e; } pair<int, point> operator &(const line &b)const { point res = s; if(sgn((s - e) ^ (b.s - b.e)) == 0) { if(sgn((s - b.e) ^ (b.s - b.e)) == 0) return make_pair(0, res); //重合 else return make_pair(1, res); //平行 } long double t = ((s - b.s) ^ (b.s - b.e)) / ((s - e) ^ (b.s - b.e)); res.x += (e.x - s.x) * t; res.y += (e.y - s.y) * t; return make_pair(2, res); } }; point lst[maxn]; int stc[maxn],top; bool _cmp(point p1,point p2) { double temp=(p1-lst[0])^(p2-lst[0]); if(sgn(temp)>0) return true; else if(sgn(temp)==0&&sgn(dist(p1,lst[0])-dist(p2,lst[0]))<=0) return true; else return false; } bool on_line(point p,line uu) { return (sgn(p.x-uu.s.x)*sgn(p.x-uu.e.x))<=0&&(sgn(p.y-uu.s.y)*sgn(p.y-uu.e.y)<=0); } void graham() { if(n==0) { top=0; return; } point p0=lst[0]; int k=0; for(int i=1;i<n;i++) { if((p0.y>lst[i].y)||(p0.y==lst[i].y&&p0.x>lst[i].x)) { p0=lst[i]; k=i; } } swap(lst[k],lst[0]); sort(lst+1,lst+n,_cmp); if(n==1) { top=1; stc[0]=0; return ; } top=2; stc[0]=0; stc[1]=1; if(n==2) return ; for(int i=2;i<n;i++) { while(top>1&&sgn((lst[stc[top-1]]-lst[stc[top-2]])^(lst[i]-lst[stc[top-2]]))<=0) top--; stc[top++]=i; } } bool in_tri(point p) { double s=fabs((tri[0]-tri[1])^(tri[0]-tri[2])); double s1=fabs((p-tri[0])^(p-tri[1])); double s2=fabs((p-tri[1])^(p-tri[2])); double s3=fabs((p-tri[2])^(p-tri[0])); return sgn(s1+s2+s3-s)==0; } bool in_rec(point p) { return sgn(p.x-xx[3])>=0&&sgn(p.x-xx[4])<=0&&sgn(p.y-yy[3])>=0&&sgn(p.y-yy[4])<=0; } int main() { #ifndef ONLINE_JUDGE freopen("in.txt","r",stdin); #endif while(~scanf("%lf%lf%lf%lf%lf%lf%lf%lf",xx+1,yy+1,xx+2,yy+2,xx+3,yy+3,xx+4,yy+4)) { tri[0]=point(xx[1],yy[1]),tri[1]=point(xx[1],yy[2]),tri[2]=point(xx[2],yy[1]); rec[0]=point(xx[3],yy[3]),rec[1]=point(xx[3],yy[4]),rec[2]=point(xx[4],yy[4]),rec[3]=point(xx[4],yy[3]); n=0; rep(i,0,3) if(in_tri(rec[i])) lst[n++]=rec[i]; rep(i,0,2) if(in_rec(tri[i])) lst[n++]=tri[i]; rep(i,0,3) { line gg=line(rec[i],rec[(i+1)%4]); rep(j,0,2) { line hh=line(tri[j],tri[(j+1)%3]); pair<int, point> res=hh≫ if(res.first==2) { point &uu=res.second; if(on_line(uu,gg)&&on_line(uu,hh)) lst[n++]=uu; } } } sort(lst,lst+n,_cmp); n=unique(lst,lst+n)-lst; graham(); double ans=0; if(top>=3) { rep(i,0,top-1) { ans+=lst[stc[i]]^lst[stc[(i+1)%top]]; } } ans/=2; printf("%.7f\n",ans); } }
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原文地址：http://www.cnblogs.com/youmi/p/5877578.html
