9.7联合作战战果

1.处理内容

数据结构部

线段树 1题

树状数组 1题

数学几何部

凸包 1题

旋转卡壳 1题

半平面交 1题

几何基础 1题

皮克定理 1题

平面分治 1题

2.数据结构

（1）线段树练习3

区间修改http://codevs.cn/problem/1082/

#include<iostream>#include<cstdio>#include<cstdlib>#include<cstring>#include<string>#include<ctime>#include<cmath>#include<algorithm>#include<cctype>#include<iomanip>#define int long longusing namespace std;inline int read(){int i=0,f=1;char ch;for(ch=getchar();!isdigit(ch);ch=getchar())if(ch=='-') f=-1;for(;isdigit(ch);ch=getchar())i=(i<<1)+(i<<3)+(ch^48);return i*f;}int buf[1024];inline void write(int x){if(!x){putchar('0');return ;}if(x<0){putchar('-');x=-x;}while(x){buf[++buf[0]]=x%10,x/=10;}while(buf[0]) putchar(48+buf[buf[0]--]);return ;}#define stan 222222int n,q,o,a,b,c,val[stan],tag[stan*4],sum[stan*4];void build(int k,int l,int r){if(l==r){sum[k]=val[l];return ;}int mid=l+r>>1;build(k<<1,l,mid);build(k<<1|1,mid+1,r);sum[k]=sum[k<<1]+sum[k<<1|1];return ;}void pushdown(int x,int l,int r){int mid=l+r>>1;tag[x<<1]+=tag[x];tag[x<<1|1]+=tag[x];sum[x<<1]+=tag[x]*(mid-l+1);sum[x<<1|1]+=tag[x]*(r-mid);tag[x]=0;return ;}void update(int k,int l,int r,int x,int y,int c){if(x<=l&&r<=y){tag[k]+=c;sum[k]+=(r-l+1)*c;return ;}if(tag[k])pushdown(k,l,r);int mid=l+r>>1;if(x<=mid) update(k<<1,l,mid,x,y,c);if(y>mid) update(k<<1|1,mid+1,r,x,y,c);sum[k]=sum[k<<1]+sum[k<<1|1];return ;}int query(int k,int l,int r,int x,int y){if(x<=l&&r<=y) return sum[k];if(tag[k]) pushdown(k,l,r);int mid=l+r>>1,ret=0;if(x<=mid) ret+=query(k<<1,l,mid,x,y);if(y>mid) ret+=query(k<<1|1,mid+1,r,x,y);return ret;}signed main(){n=read();for(int i=1;i<=n;++i)val[i]=read();build(1,1,n);q=read();for(int i=1;i<=q;++i){o=read();if(o==1){a=read();b=read();c=read();update(1,1,n,a,b,c);}else{a=read();b=read();if(a>b) swap(a,b);write(query(1,1,n,a,b));puts(" ");}}return 0;}

（2）线段树练习2

可以直接上树状数组http://codevs.cn/problem/1081/

#include<iostream>#include<cstdio>#include<cstdlib>#include<cstring>#include<string>#include<ctime>#include<cmath>#include<algorithm>#include<cctype>#include<iomanip>#define int long longusing namespace std;inline int read(){int i=0,f=1;char ch;for(ch=getchar();!isdigit(ch);ch=getchar())if(ch=='-') f=-1;for(;isdigit(ch);ch=getchar())i=(i<<1)+(i<<3)+(ch^48);return i*f;}int buf[1024];inline void write(int x){if(!x){putchar('0');return ;}if(x<0){putchar('-');x=-x;}while(x){buf[++buf[0]]=x%10,x/=10;}while(buf[0]) putchar(48+buf[buf[0]--]);return ;}#define stan 111111int n,q,o,a,b,c,val[stan],data[stan];void insert(int x,int v){while(x<=n){data[x]+=v;x+=(x&(-x));}return ;}int query(int x){int ret=0;while(x){ret+=data[x];x-=(x&(-x));}return ret;}signed main(){n=read();for(int i=1;i<=n;++i){val[i]=read();insert(i,val[i]);}q=read();for(int i=1;i<=q;++i){o=read();if(o==1){a=read();b=read();c=read();for(int j=a;j<=b;++j)insert(j,c);}else{a=read();write(query(a)-query(a-1));puts(" ");}}return 0;}

3.数学几何部

（1）圈奶牛

求最大凸包周长http://www.cogs.pro/cogs/problem/problem.php?pid=896

提供两种极角排序算法

algorithm1.graham-scan

#include<iostream>#include<cstdio>#include<cstdlib>#include<cstring>#include<string>#include<ctime>#include<cmath>#include<algorithm>#include<cctype>#include<iomanip>#define int long long#define eps 1e-7using namespace std;inline int read(){int i=0,f=1;char ch;for(ch=getchar();!isdigit(ch);ch=getchar())if(ch=='-') f=-1;for(;isdigit(ch);ch=getchar())i=(i<<1)+(i<<3)+(ch^48);return i*f;}int buf[1024];inline void write(int x){if(!x){putchar('0');return ;}if(x<0){putchar('-');x=-x;}while(x){buf[++buf[0]]=x%10,x/=10;}while(buf[0]) putchar(48+buf[buf[0]--]);return ;}#define stan 11111struct P{double x,y;}p[stan],q[stan];int n,top,order[stan]; double ans;P vec(const P &a,const P &b){P q;q.x=a.x-b.x;q.y=a.y-b.y;return q;}double xplus(P a,P b){return a.x*b.y-a.y*b.x;}double dotplus(P a){return a.x*a.x+a.y*a.y;}bool cmp(int a,int b){int x=xplus(vec(p[a],p[1]),vec(p[b],p[1]));if(x!=0) return x>0;else return dotplus(vec(p[a],p[1]))<dotplus(vec(p[b],p[1]));}double dist(P a,P b){return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));}void graham(){top=1;for(int i=2;i<=n;++i)if(p[i].x<p[top].x||p[i].x==p[top].x&&p[i].y<p[top].y)top=i;if(top!=1)swap(p[1],p[top]);for(int i=1;i<=n;++i)order[i]=i;sort(order+1,order+n+1,cmp);top=0;q[++top]=p[order[1]];for(int i=2;i<=n;++i){int u=order[i];while(top>1&&xplus(vec(p[u],q[top-1]),vec(q[top],q[top-1]))>=0) --top;q[++top]=p[u];}for(int i=1;i<top;++i)ans+=dist(q[i],q[i+1]);if(top>2) ans+=dist(q[1],q[top]);return ;}signed main(){//freopen("fc.in","r",stdin);//freopen("fc.out","w",stdout);n=read();for(int i=1;i<=n;++i)scanf("%lf%lf",&p[i].x,&p[i].y);graham();printf("%.2lf",ans);return 0;}

algorithm2.鸡肋的melkman

据说是在线，但并不会用在线版

思想:在当前点有序的情况下，维护一个支持两端插入/删除的队列

总觉得和graham没什么区别...

wuvin大佬也是这么觉得的....

#include<iostream>#include<cstdio>#include<cstdlib>#include<cstring>#include<string>#include<ctime>#include<cmath>#include<algorithm>#include<cctype>#include<iomanip>#define int long long#define eps 1e-7using namespace std;inline int read(){int i=0,f=1;char ch;for(ch=getchar();!isdigit(ch);ch=getchar())if(ch=='-') f=-1;for(;isdigit(ch);ch=getchar())i=(i<<1)+(i<<3)+(ch^48);return i*f;}int buf[1024];inline void write(int x){if(!x){putchar('0');return ;}if(x<0){putchar('-');x=-x;}while(x){buf[++buf[0]]=x%10,x/=10;}while(buf[0]) putchar(48+buf[buf[0]--]);return ;}#define stan 11111struct P{double x,y;}p[stan],q[stan*2];int n,top,order[stan],guideb,guidet; double ans;P vec(const P &a,const P &b){P q;q.x=a.x-b.x;q.y=a.y-b.y;return q;}double xplus(P a,P b){return a.x*b.y-a.y*b.x;}double dotplus(P a){return a.x*a.x+a.y*a.y;}bool cmp(int a,int b){int x=xplus(vec(p[a],p[1]),vec(p[b],p[1]));if(x!=0) return x>0;else return dotplus(vec(p[a],p[1]))<dotplus(vec(p[b],p[1]));}double dist(P a,P b){return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));}void melkman(){top=1;for(int i=2;i<=n;++i)if(p[i].x<p[top].x||p[i].x==p[top].x&&p[i].y<p[top].y)top=i;if(top!=1)swap(p[1],p[top]);for(int i=1;i<=n;++i)order[i]=i;sort(order+1,order+n+1,cmp);guideb=guidet=n;q[n]=p[order[1]];q[++guidet]=p[order[2]];q[++guidet]=p[order[3]];q[--guideb]=p[order[3]];for(int i=4;i<=n;++i){int u=order[i];while(guidet-guideb>1&&xplus(vec(p[u],q[guidet-1]),vec(q[guidet],q[guidet-1]))>=eps) --guidet;q[++guidet]=p[u];while(guidet-guideb>1&&xplus(vec(q[guideb],q[guideb+1]),vec(p[u],q[guideb+1]))>=-eps) ++guideb;q[--guideb]=p[u];}for(int i=guideb;i<guidet;++i)ans+=dist(q[i],q[i+1]);return ;}signed main(){freopen("fc.in","r",stdin);freopen("fc.out","w",stdout);n=read();for(int i=1;i<=n;++i)scanf("%lf%lf",&p[i].x,&p[i].y);melkman();printf("%.2lf",ans);return 0;}

（2）最大土地面积(SCOI2007)

在凸包的基础上枚举对角线http://www.lydsy.com/JudgeOnline/problem.php?id=1069

同样提供两种极角排序的凸包算法

#include<iostream>#include<cstdio>#include<cstdlib>#include<cstring>#include<string>#include<ctime>#include<cmath>#include<algorithm>#include<cctype>#include<iomanip>#define int long long#define eps 1e-7using namespace std;inline int read(){int i=0,f=1;char ch;for(ch=getchar();!isdigit(ch);ch=getchar())if(ch=='-') f=-1;for(;isdigit(ch);ch=getchar())i=(i<<1)+(i<<3)+(ch^48);return i*f;}int buf[1024];inline void write(int x){if(!x){putchar('0');return ;}if(x<0){putchar('-');x=-x;}while(x){buf[++buf[0]]=x%10,x/=10;}while(buf[0]) putchar(48+buf[buf[0]--]);return ;}#define stan 11111struct P{double x,y;}p[stan],q[stan*2],r[stan];int n,top,order[stan],guideb,guidet; double ans;P vec(const P &a,const P &b){P q;q.x=a.x-b.x;q.y=a.y-b.y;return q;}double xplus(P a,P b){return a.x*b.y-a.y*b.x;}double dotplus(P a){return a.x*a.x+a.y*a.y;}bool cmp(int a,int b){double x=xplus(vec(p[a],p[1]),vec(p[b],p[1]));if(x!=0) return x>0;else return dotplus(vec(p[a],p[1]))<dotplus(vec(p[b],p[1]));}void melkman(){top=1;for(int i=2;i<=n;++i)if(p[i].x<p[top].x||p[i].x==p[top].x&&p[i].y<p[top].y)top=i;if(top!=1)swap(p[1],p[top]);for(int i=1;i<=n;++i)order[i]=i;sort(order+1,order+n+1,cmp);guideb=guidet=n;q[n]=p[order[1]];q[++guidet]=p[order[2]];q[++guidet]=p[order[3]];q[--guideb]=p[order[3]];for(int i=4;i<=n;++i){int u=order[i];while(guidet-guideb>1&&xplus(vec(p[u],q[guidet-1]),vec(q[guidet],q[guidet-1]))>=-eps) --guidet;q[++guidet]=p[u];while(guidet-guideb>1&&xplus(vec(q[guideb],q[guideb+1]),vec(p[u],q[guideb+1]))>=-eps) ++guideb;q[--guideb]=p[u];}top=0;for(int i=guideb;i<guidet;++i)r[++top]=q[i];return ;}void graham(){top=1;for(int i=2;i<=n;++i)if(p[i].x<p[top].x||p[i].x==p[top].x&&p[i].y<p[top].y)top=i;if(top!=1)swap(p[1],p[top]);for(int i=1;i<=n;++i)order[i]=i;sort(order+1,order+n+1,cmp);top=0;q[++top]=p[order[1]];for(int i=2;i<=n;++i){int u=order[i];while(top>1&&xplus(vec(p[u],q[top-1]),vec(q[top],q[top-1]))>=0) --top;q[++top]=p[u];}for(int i=1;i<=top;++i)r[i]=q[i];return ;}int ahead(int x){if(x==top) return 1;else return x+1;}void solve(){ans=0.000;for(int i=1;i<=top;++i){double s1,s2;for(int q1=ahead(i),j=ahead(q1),q2=ahead(j);j<top;j=ahead(j)){while(xplus(vec(r[j],r[i]),vec(r[q1],r[i]))<xplus(vec(r[j],r[i]),vec(r[ahead(q1)],r[i]))) q1=ahead(q1);s1=xplus(vec(r[j],r[i]),vec(r[q1],r[i]));while(xplus(vec(r[q2],r[i]),vec(r[j],r[i]))<xplus(vec(r[ahead(q2)],r[i]),vec(r[j],r[i]))) q2=ahead(q2);s2=xplus(vec(r[q2],r[i]),vec(r[j],r[i]));ans=max(ans,s1+s2);}}return ;}signed main(){n=read();for(int i=1;i<=n;++i)scanf("%lf%lf",&p[i].x,&p[i].y);melkman();solve();printf("%.3lf",ans*0.5);return 0;}

（3）瞭望塔(ZJOI2008)

著名的半平面交样板

甩链接吧还是http://www.lydsy.com/JudgeOnline/problem.php?id=1038

感兴趣者可以再拿byvoid大佬的大灾变做一做找自信https://www.byvoid.com/zhs/blog/cataclysm-problem

#include<iostream>#include<cstdio>#include<cstdlib>#include<cstring>#include<string>#include<ctime>#include<cmath>#include<algorithm>#include<cctype>#include<iomanip>#define int long long#define double long double#define eps 1e-7using namespace std;inline int read(){int i=0,f=1;char ch;for(ch=getchar();!isdigit(ch);ch=getchar())if(ch=='-') f=-1;for(;isdigit(ch);ch=getchar())i=(i<<1)+(i<<3)+(ch^48);return i*f;}int buf[1024];inline void write(int x){if(!x){putchar('0');return ;}if(x<0){putchar('-');x=-x;}while(x){buf[++buf[0]]=x%10,x/=10;}while(buf[0]) putchar(48+buf[buf[0]--]);return ;}#define stan 3333struct P{double x,y;}p[stan],a[stan];int n,cnt,tot; double ans;P vec(const P &a,const P &b){P q;q.x=a.x-b.x;q.y=a.y-b.y;return q;}double xplus(P a,P b){return a.x*b.y-a.y*b.x;}double dotplus(P a){return a.x*a.x+a.y*a.y;}struct L{P a,b;double slope;friend bool operator <(L x,L y){if(x.slope!=y.slope) return x.slope<y.slope;else return xplus(vec(x.b,x.a),vec(y.b,x.a))>0;} }l[stan],que[stan];P inter(L x,L y){double k1,k2,t;k1=xplus(vec(y.b,x.a),vec(x.b,x.a));k2=xplus(vec(x.b,x.a),vec(y.a,x.a));t=k1/(k1+k2);P ret;ret.x=y.b.x+t*(y.a.x-y.b.x);ret.y=y.b.y+t*(y.a.y-y.b.y);return ret;}bool check(L a,L b,L t){P s=inter(a,b);return xplus(vec(t.b,t.a),vec(s,t.a))<0;}void preact(){p[0].x=p[1].x;p[0].y=999999;p[n+1].x=p[n].x;p[n+1].y=999999;for(int i=1;i<=n;++i){l[++cnt].a=p[i-1];l[cnt].b=p[i];l[++cnt].a=p[i];l[cnt].b=p[i+1];}for(int i=1;i<=cnt;++i)l[i].slope=atan2(l[i].b.y-l[i].a.y,l[i].b.x-l[i].a.x);sort(l+1,l+cnt+1);return ;}void hpi(){int ll=1,rr=0;tot=0;for(int i=1;i<=cnt;++i){if(l[i].slope!=l[i-1].slope) ++tot;l[tot]=l[i];}cnt=tot,tot=0;que[++rr]=l[1];que[++rr]=l[2];for(int i=3;i<=cnt;++i){while(ll<rr&&check(que[rr-1],que[rr],l[i])) --rr;while(ll<rr&&check(que[ll+1],que[ll],l[i])) ++ll;que[++rr]=l[i];}while(ll<rr&&check(que[rr-1],que[rr],que[ll])) --rr;while(ll<rr&&check(que[ll+1],que[ll],que[rr])) ++ll;for(int i=ll;i<rr;++i)a[++tot]=inter(que[i],que[i+1]);return ;}void getans(){for(int i=1;i<=tot;++i){P u;u.x=a[i].x;u.y=-1;for(int j=1;j<n;++j)if(a[i].x>=p[j].x&&a[i].x<=p[j+1].x)ans=min(ans,a[i].y-inter((L){p[j],p[j+1]},(L){u,a[i]}).y);}for(int i=1;i<=n;++i){P u;u.x=p[i].x;u.y=-1;for(int j=1;j<tot;++j)if(p[i].x>=a[j].x&&p[i].x<=a[j+1].x)ans=min(ans,inter((L){a[j],a[j+1]},(L){u,p[i]}).y-p[i].y);}return ;}signed main(){ans=1e60;n=read();for(int i=1;i<=n;++i)p[i].x=read();for(int i=1;i<=n;++i)p[i].y=read();preact();hpi();getans();printf("%.3Lf",ans);return 0;}

（4）segments

这个题反正可以转化为求与所有线段相交的直线http://poj.org/problem?id=3304

至于复杂度，O(n^3)可以轻松过100

怎么判?用差乘判

#include<iostream>#include<cstdio>#include<cstdlib>#include<cstring>#include<string>#include<ctime>#include<cmath>#include<algorithm>#include<cctype>#include<iomanip>#define eps 1e-8using namespace std;inline int read(){int i=0,f=1;char ch;for(ch=getchar();!isdigit(ch);ch=getchar())if(ch=='-') f=-1;for(;isdigit(ch);ch=getchar())i=(i<<1)+(i<<3)+(ch^48);return i*f;}int buf[1024];inline void write(int x){if(!x){putchar('0');return ;}if(x<0){putchar('-');x=-x;}while(x){buf[++buf[0]]=x%10,x/=10;}while(buf[0]) putchar(48+buf[buf[0]--]);return ;}#define stan 1111111struct P{double x,y;}p[stan];int n,cnt,tot,T,tag; double ans;P vec(const P &a,const P &b){P q;q.x=a.x-b.x;q.y=a.y-b.y;return q;}double xplus(P a,P b){return a.x*b.y-a.y*b.x;}double dotplus(P a){return a.x*a.x+a.y*a.y;}struct L{P a,b;double slope;friend bool operator <(L x,L y){if(x.slope!=y.slope) return x.slope<y.slope;else return xplus(vec(x.b,x.a),vec(y.b,x.a))>0;} }l[stan],que[stan];P inter(L x,L y){double k1,k2,t;k1=xplus(vec(y.b,x.a),vec(x.b,x.a));k2=xplus(vec(x.b,x.a),vec(y.a,x.a));t=k1/(k1+k2);P ret;ret.x=y.b.x+t*(y.a.x-y.b.x);ret.y=y.b.y+t*(y.a.y-y.b.y);return ret;}bool check(P a,P b){if(fabs(a.x-b.x)<eps&&fabs(a.y-b.y)<eps) return false;for(int i=1;i<=n;++i)if(xplus(vec(l[i].a,a),vec(b,a))*xplus(vec(l[i].b,a),vec(b,a))>eps) return false;return true;}signed main(){T=read();while(T--){n=read();for(int i=1;i<=n;++i)scanf("%lf%lf%lf%lf",&l[i].a.x,&l[i].a.y,&l[i].b.x,&l[i].b.y);tag=true;if(n<3) tag=false;for(int i=1;i<=n&&tag;++i){if(check(l[i].a,l[i].b)) tag=false;for(int j=i+1;j<=n&&tag;++j){if(check(l[i].a,l[j].a)) tag=false;else if(check(l[i].b,l[j].a)) tag=false;else if(check(l[i].a,l[j].b)) tag=false;else if(check(l[i].b,l[j].b)) tag=false;}}if(tag)puts("No!");elseputs("Yes!");}return 0;}

（5）Triangle

一个多边形的顶点如果全是格点

则有 2S=2a+b-2

其中S是多边形面积，a是多边形内部格点数，b是多边形上的格点数

至于如何求S，差乘即可

至于如何求b，求gcd即可

#include<iostream>#include<cstdio>#include<cstdlib>#include<cstring>#include<string>#include<ctime>#include<cmath>#include<algorithm>#include<cctype>#include<iomanip>using namespace std;inline int read(){int i=0,f=1;char ch;for(ch=getchar();!isdigit(ch);ch=getchar())if(ch=='-') f=-1;for(;isdigit(ch);ch=getchar())i=(i<<1)+(i<<3)+(ch^48);return i*f;}int buf[1024];inline void write(int x){if(!x){putchar('0');return ;}if(x<0){putchar('-');x=-x;}while(x){buf[++buf[0]]=x%10,x/=10;}while(buf[0]) putchar(48+buf[buf[0]--]);return ;}int a,b,c,d,e,f,B,S;int gcd(int a,int b){if(b) return gcd(b,a%b);return a;}signed main(){a=read();b=read();c=read();d=read();e=read();f=read();while(a||b||c||d||e||f){B=0;S=0;B+=gcd(abs(a-c),abs(b-d));B+=gcd(abs(c-e),abs(d-f));B+=gcd(abs(e-a),abs(f-b));S=abs((c-a)*(f-b)-(e-a)*(d-b));write((S+2-B)/2);puts(" ");a=read();b=read();c=read();d=read();e=read();f=read();}}

（6）Quoit Design

很神奇的O(nlogn)分治

先预处理两块之间的分界线，递归的结果在x轴框一个范围

然后对于合法点，再在y轴框一个范围

处理之后递归

#include<iostream>#include<cstdio>#include<cstdlib>#include<cstring>#include<string>#include<ctime>#include<cmath>#include<algorithm>#include<cctype>#include<iomanip>#define eps 1e-7using namespace std;inline int read(){int i=0,f=1;char ch;for(ch=getchar();!isdigit(ch);ch=getchar())if(ch=='-') f=-1;for(;isdigit(ch);ch=getchar())i=(i<<1)+(i<<3)+(ch^48);return i*f;}int buf[1024];inline void write(int x){if(!x){putchar('0');return ;}if(x<0){putchar('-');x=-x;}while(x){buf[++buf[0]]=x%10,x/=10;}while(buf[0]) putchar(48+buf[buf[0]--]);return ;}#define stan 1111111struct P{double x,y;}p[stan];int n,cnt,tot; double ans;P vec(const P &a,const P &b){P q;q.x=a.x-b.x;q.y=a.y-b.y;return q;}double xplus(P a,P b){return a.x*b.y-a.y*b.x;}double dotplus(P a){return a.x*a.x+a.y*a.y;}bool cmp(P a,P b){return a.x<b.x;}double dist(P a,P b){return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));}double solve(int l,int r){if(l==r) return 1e20;int mid=l+r>>1;double lx=(p[mid].x+p[mid+1].x)/2.0;double dis=min(solve(l,mid),solve(mid+1,r));static P pre[stan],suf[stan],tmp[stan];int x=l,y=mid+1,nop=0,nos=0;for(int i=l;i<=r;++i){if(x<=mid&&(y==r+1||p[x].y<p[y].y)){tmp[i]=p[x];++x;if(lx-dis<tmp[i].x)pre[++nop]=tmp[i];}else{tmp[i]=p[y];++y;if(lx+dis>tmp[i].x)suf[++nos]=tmp[i];}}for(int i=l;i<=r;++i)p[i]=tmp[i];for(int i=1,j=1;i<=nop||j<=nos;){if(i<=nop&&(j>nos||pre[i].y<suf[j].y)){for(int k=j-1;k;--k){if(pre[i].y-dis>=suf[k].y) break;dis=min(dis,dist(pre[i],suf[k]));}++i;}else{for(int k=i-1;k;--k){if(suf[j].y-dis>=pre[k].y) break;dis=min(dis,dist(pre[k],suf[j]));}++j;}}return dis;}signed main(){while(n=read(),n!=0){for(int i=1;i<=n;++i)scanf("%lf%lf",&p[i].x,&p[i].y);sort(p+1,p+n+1,cmp);printf("%.2lf\n",solve(1,n)/2.0);}return 0;}

局部完结撒花