BZOJ3839: [Pa2013]Działka

来源：互联网发布：建筑设计优化的方法编辑：程序博客网时间：2024/05/01 10:32
卡时限过去
~~(多刷几次就好啦~~
凸壳分成四块扫描线加线段树处理边界点
根据预处理的边界点之间凸壳面积计算
面积计算以原点为三角形的第三个节点注意每一块面积对答案的贡献是正的还是负的
~~这题居然花了我一天时间~~
#include<cstdio>#include<iostream>#include<cstring>#include<algorithm>using namespace std;#define abs(a) ((a)<0?-(a):(a))#define ll long longll A[3001][3001],B[3001][3001],C[3001][3001],D[3001][3001];struct Flag{    int a,b;    Flag(){};    Flag(int a,int b):a(a),b(b){}    inline friend Flag operator +(Flag a,Flag b){return a.b>b.b?a:b;}    inline friend bool operator !=(Flag a,Flag b){return a.b^b.b;}};Flag NUL=Flag(-1,-1);struct Node{    Flag A,f;    int l,r;    Node *lc,*rc;    Node():f(NUL),A(NUL),lc(NULL),rc(NULL){};    Node(int x,int y):l(x),r(y),A(NUL),lc(NULL),rc(NULL),f(NUL){};}Q7[1000001];int Cnt;inline Node *Build(int l,int r){Q7[++Cnt]=Node(l,r);return Q7+Cnt;}inline void pushdown(Node *Cur){    if(Cur->lc==NULL)Cur->lc=Build(Cur->l,Cur->l+Cur->r>>1),Cur->rc=Build((Cur->l+Cur->r>>1)+1,Cur->r);    Cur->lc->A=Cur->rc->A=Cur->lc->f=Cur->rc->f=Cur->f;    Cur->f=NUL;}inline void Update(Node *Cur,int l,Flag Delt){    if(!Cur)return;    if(Cur->l>=l&&l>=Cur->r){Cur->A=Cur->f=Delt;return;}    if(Cur->f!=NUL)pushdown(Cur);    Cur->A=Delt;    int mid=Cur->l+Cur->r>>1;    if(!Cur->lc)        Cur->lc=Build(Cur->l,mid),Cur->rc=Build(mid+1,Cur->r);    if(l<=mid)Update(Cur->lc,l,Delt);    else Update(Cur->rc,l,Delt);}Flag Query(Node *Cur,int l,int r){    if(!Cur)return NUL;    int mid=Cur->l+Cur->r>>1;    if(l<=Cur->l&&r>=Cur->r)return Cur->A;    Flag res=NUL;    if(l<=mid)res=res+Query(Cur->lc,l,r);    if(r>mid)res=res+Query(Cur->rc,l,r);    return res+Cur->f;}struct P{    int x,y,i;    P(){}    P(int x,int y):x(x),y(y){}    inline friend ll operator *(P a,P b){return a.x*1ll*b.y-a.y*1ll*b.x;}    inline friend P operator -(P a,P b){return P(a.x-b.x,a.y-b.y);}}L[3001];inline bool cmp1(P a,P b){return a.x^b.x?a.x<b.x:a.y<b.y;}inline bool cmp2(P a,P b){return a.x^b.x?a.x<b.x:a.y>b.y;}inline bool cmp3(P a,P b){return a.x^b.x?a.x>b.x:a.y<b.y;}inline bool cmp4(P a,P b){return a.x^b.x?a.x>b.x:a.y>b.y;}inline bool cmp5(P a,P b){return a.x^b.x?a.x<b.x:a.y<b.y;}inline bool cmp6(P a,P b){return a.y^b.y?a.y<b.y:a.x<b.x;}inline bool cmp7(P a,P b){return a.x^b.x?a.x>b.x:a.y>b.y;}inline bool cmp8(P a,P b){return a.y^b.y?a.y>b.y:a.x>b.x;}struct Q{    int P;    int no;}sx[1000001],ex[1000001],sy[1000001],ey[1000001];int EX[1000001],EY[1000001],SX[1000001],SY[1000001];int  Ex[1000001],Ey[1000001],Sx[1000001],Sy[1000001];bool Ans[1000001];inline bool cmps(Q a,Q b){return a.P<b.P;}inline bool cmpe(Q a,Q b){return a.P>b.P;}P Bl=P(-0,-0);char c;bool flag;inline void read(int &a){a=0;do c=getchar(); while(c!='-'&&(c<'0'||c>'9'));c=c=='-'?flag=true,getchar():c;while(c<='9'&&c>='0')a=(a<<3)+(a<<1)+c-'0',c=getchar();a=flag?flag=false,-a:a;}int stack[100001],top;int main(){    int n,k;    read(k),read(n);    for(int i=1;i<=n;i++)read(L[i].x),read(L[i].y),L[i].i=i;    sort(L+1,L+1+n,cmp1);    for(int i=1;i<n;i++)        {            stack[top=1]=i;            for(int j=i+1;j<=n;j++)if(L[j].y>=L[i].y)            {                while(top>1&&(L[stack[top]]-L[stack[top-1]])*(L[j]-L[stack[top-1]])<=0)top--;                stack[++top]=j;                A[L[i].i][L[j].i]=L[stack[top-1]]*L[stack[top]]+A[L[i].i][L[stack[top-1]].i];            }        }    sort(L+1,L+1+n,cmp2);    for(int i=1;i<n;i++)        {            stack[top=1]=i;            for(int j=i+1;j<=n;j++)if(L[j].y<=L[i].y)            {                while(top>1&&(L[stack[top]]-L[stack[top-1]])*(L[j]-L[stack[top-1]])>=0)top--;                stack[++top]=j;                B[L[i].i][L[j].i]=L[stack[top-1]]*L[stack[top]]+B[L[i].i][L[stack[top-1]].i];            }        }    sort(L+1,L+1+n,cmp3);    for(int i=1;i<n;i++)        {            stack[top=1]=i;             for(int j=i+1;j<=n;j++)if(L[j].y>=L[i].y)            {                while(top>1&&(L[stack[top]]-L[stack[top-1]])*(L[j]-L[stack[top-1]])>=0)top--;                stack[++top]=j;                C[L[i].i][L[j].i]=L[stack[top-1]]*L[stack[top]]+C[L[i].i][L[stack[top-1]].i];            }        }    sort(L+1,L+1+n,cmp4);    for(int i=1;i<n;i++)        {            stack[top=1]=i;            for(int j=i+1;j<=n;j++)if(L[j].y<=L[i].y)            {                while(top>1&&(L[stack[top]]-L[stack[top-1]])*(L[j]-L[stack[top-1]])<=0)top--;                stack[++top]=j;                D[L[i].i][L[j].i]=L[stack[top-1]]*L[stack[top]]+D[L[i].i][L[stack[top-1]].i];            }        }    int m;    read(m);    for(int i=1;i<=m;i++)        read(sx[i].P),read(ex[i].P),read(sy[i].P),read(ey[i].P),        sx[i].no=sy[i].no=ex[i].no=ey[i].no=i,        Sx[i]=sx[i].P,Sy[i]=sy[i].P,Ex[i]=ex[i].P,Ey[i]=ey[i].P;    sort(L+1,L+1+n,cmp6);    sort(ey+1,ey+1+m,cmps);    int j=1;    Cnt=0;    Node *MYT=Build(0,k);    for(int i=1;i<=m;i++)    {        while(j<=n&&L[j].y<=ey[i].P)Update(MYT,L[j].x,Flag(L[j].i,j)),j++;        Flag T;        EY[ey[i].no]=(T=Query(MYT,Sx[ey[i].no],Ex[ey[i].no])).a;        if(T.b==-1||Sy[ey[i].no]>L[T.b].y)Ans[ey[i].no]=true;    }Cnt=0;    MYT=Build(0,k);    sort(L+1,L+1+n,cmp8);    sort(sy+1,sy+1+m,cmpe);    j=1;    for(int i=1;i<=m;i++)    {        while(j<=n&&L[j].y>=sy[i].P)Update(MYT,L[j].x,Flag(L[j].i,j)),j++;        Flag T;        SY[sy[i].no]=(T=Query(MYT,Sx[sy[i].no],Ex[sy[i].no])).a;        if(T.b==-1||Ey[sy[i].no]<L[T.b].y)Ans[sy[i].no]=true;    }Cnt=0;    MYT=Build(0,k);    sort(L+1,L+1+n,cmp5);    sort(ex+1,ex+1+m,cmps);    j=1;    for(int i=1;i<=m;i++)    {        while(j<=n&&L[j].x<=ex[i].P)Update(MYT,L[j].y,Flag(L[j].i,j)),j++;        Flag T;        EX[ex[i].no]=(T=Query(MYT,Sy[ex[i].no],Ey[ex[i].no])).a;        if(T.b==-1||Sx[ex[i].no]>L[T.b].x)Ans[ex[i].no]=true;    }    sort(L+1,L+1+n,cmp7);    sort(sx+1,sx+1+m,cmpe);    j=1;Cnt=0;    MYT=Build(0,k);    for(int i=1;i<=m;i++)    {        while(j<=n&&L[j].x>=sx[i].P)Update(MYT,L[j].y,Flag(L[j].i,j)),j++;        Flag T;        SX[sx[i].no]=(T=Query(MYT,Sy[sx[i].no],Ey[sx[i].no])).a;        if(T.b==-1||Ex[sx[i].no]<L[T.b].x)Ans[sx[i].no]=true;    }    for(int i=1;i<=m;i++)    {        ll ans=0;        if(SY[i]==-1||EY[i]==-1||EX[i]==-1||SX[i]==-1){puts("0.0");continue;}        if(Ans[i]){puts("0.0");continue;}        ans=A[SY[i]][EX[i]]-B[EY[i]][EX[i]]-C[SY[i]][SX[i]]+D[EY[i]][SX[i]];        printf("%lld.%d\n",ans>>1,ans&1?5:0);    }    return 0;}
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