分治分块与计算几何练习 [Cloned]
https://cn.vjudge.net/contest/148706
A
#include<cstdio> #include<cstring> #include<cmath> #include<iostream> #include<algorithm> using namespace std; typedef long long ll; int read(){ int x=0,f=1;char ch=getchar(); while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();} while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();} return x*f; } const int N=2e5+7; struct node{ int l,r,t,pos; bool operator < (const node &a)const{ return pos==a.pos?r<a.r:pos<a.pos; } }b[N]; int n,m,l,r,a[N],f[N]; ll res,ans1[N],ans2[N]; ll gcd(ll a,ll b){ if(!b) return a; return gcd(b,a%b); } int main(){ n=read();m=read(); for(int i=1;i<=n;i++) a[i]=read(); int k=sqrt(n*1.0)+0.5; for(int i=1;i<=m;i++){ b[i].l=read();b[i].r=read(); b[i].t=i;b[i].pos=b[i].l/k; } sort(b+1,b+m+1); memset(f,0,sizeof f); l=1;r=0;res=0; for(int i=1;i<=m;i++){ while(r>b[i].r){ res-=(ll)f[a[r]]-1; f[a[r]]--; r--; } while(r<b[i].r){ r++; f[a[r]]++; res+=(ll)f[a[r]]-1; } while(l>b[i].l){ l--; f[a[l]]++; res+=(ll)f[a[l]]-1; } while(l<b[i].l){ res-=(ll)f[a[l]]-1; f[a[l]]--; l++; } ans1[b[i].t]=res; ans2[b[i].t]=(ll)(r-l+1)*(r-l)/2; } for(int i=1;i<=m;i++){ if(!ans1[i]){ puts("0/1"); continue; } ll gg=gcd(ans1[i],ans2[i]); printf("%lld/%lld\n",ans1[i]/gg,ans2[i]/gg); //printf("%I64d/%I64d\n",ans1[i]/gg,ans2[i]/gg); } return 0; }
B
//扫描线考精度，没什么技术含量 #include<cstdio> #include<cmath> #include<algorithm> #define eps 1e-8 using namespace std; int n,tot,cnt; double last,ans,pos[100005]; struct P{double x,y;}p[105][3]; struct L{P a,b;}l[105][3]; struct seg{double l,r;}f[105]; inline P operator -(P a,P b){return (P){a.x-b.x,a.y-b.y};} inline double operator *(P a,P b){return a.x*b.y-a.y*b.x;}//叉积 inline double operator /(P a,P b){return a.x*b.x+a.y*b.y;}//点积 inline P inter(L l1,L l2){ double k1=(l2.b-l1.a)*(l1.b-l1.a),k2=(l1.b-l1.a)*(l2.a-l1.a),t=k1/(k1+k2); return (P){l2.b.x+(l2.a.x-l2.b.x)*t,l2.b.y+(l2.a.y-l2.b.y)*t}; } inline bool judge(L l1,L l2){ return fabs((l1.b.y-l1.a.y)*(l2.b.x-l2.a.x)-(l1.b.x-l1.a.x)*(l2.b.y-l2.a.y))>eps; } inline bool cmp(seg a,seg b){ return fabs(a.l-b.l)<=eps?a.r<b.r:a.l<b.l; } inline double dcmp(double x){ if(fabs(x)<=eps) return 0; else return x<0?-1:1; } inline bool cross(P a1,P a2,P b1,P b2){ double c1=(a2-a1)*(b1-a1),c2=(a2-a1)*(b2-a1),c3=(b2-b1)*(a1-b1),c4=(b2-b1)*(a2-b1); return dcmp(c1)*dcmp(c2)<0&&dcmp(c3)*dcmp(c4)<0; } inline double calc(double x){ L ln=(L){(P){x,0},(P){x,1}}; int num;cnt=0; double y[4],h,ret=0; for(int i=1;i<=n;i++){ double mn=min(p[i][0].x,min(p[i][1].x,p[i][2].x)),mx=max(p[i][0].x,max(p[i][1].x,p[i][2].x)); if(x<mn+eps||x>mx-eps) continue; num=0; for(int j=0;j<=2;j++){ if(judge(l[i][j],ln)){ P tmp=inter(l[i][j],ln); if((l[i][j].a-tmp)/(l[i][j].b-tmp)>-eps) continue; y[++num]=tmp.y; } } if(num>1) f[++cnt]=(seg){y[1],y[2]}; } for(int i=1;i<=cnt;i++){ if(f[i].l>f[i].r){ swap(f[i].l,f[i].r); } } sort(f+1,f+cnt+1,cmp); for(int i=1;i<=cnt;i++){ if(i==1||f[i].l>h+eps) ret+=f[i].r-f[i].l,h=f[i].r; else if(f[i].r>h+eps) ret+=f[i].r-h,h=f[i].r; } return ret; } int main(){ scanf("%d",&n); for(int i=1;i<=n;i++){ for(int j=0;j<=2;j++){ scanf("%lf%lf",&p[i][j].x,&p[i][j].y),pos[++tot]=p[i][j].x; } } for(int i=1;i<=n;i++){ l[i][0]=(L){p[i][1],p[i][2]}, l[i][1]=(L){p[i][0],p[i][2]}, l[i][2]=(L){p[i][0],p[i][1]}; } for(int i=1;i<n;i++){ for(int j=i+1;j<=n;j++){ for(int k1=0;k1<=2;k1++){ for(int k2=0;k2<=2;k2++){ if(cross(l[i][k1].a,l[i][k1].b,l[j][k2].a,l[j][k2].b)){ pos[++tot]=inter(l[i][k1],l[j][k2]).x; } } } } } sort(pos+1,pos+tot+1); last=pos[1]; for(int i=2;i<=tot;i++){ if(fabs(pos[i]-last)>eps){ ans+=calc((pos[i]+last)/2)*(pos[i]-last); last=pos[i]; } } ans-=eps;//eps! printf("%.2lf\n",ans); return 0; }
C
//题意：n场考试中分别答对a_i题，总题数分别为b_i，允许去掉k场考试，求能达到的最高准确率。 //二分比0/1分数规划慢，然而后者并不会 #include<cstdio> #include<algorithm> #define eps 1e-7 using namespace std; const int N=1005; int n,k; double sum,a[N],b[N],t[N]; int main(){ while(scanf("%d%d",&n,&k)==2){ if(!n) break; for(int i=0;i<n;i++) scanf("%lf",&a[i]); for(int i=0;i<n;i++) scanf("%lf",&b[i]); double l=0.0,r=1.0,mid; while(r-l>eps){ mid=(l+r)/2.0; for(int i=0;i<n;i++) t[i]=a[i]-mid*b[i]; sort(t,t+n);sum=0; for(int i=k;i<n;i++) sum+=t[i]; if(sum>0) l=mid; else r=mid; } printf("%.0f\n",mid*100);//g++，坑 } return 0; }
D
//simpson自适应公式 #include<cstdio> #include<cstdlib> #include<cmath> #include<iostream> using namespace std; #define sqr(x) ((x)*(x)) typedef double DB; const int maxn=510;const DB zero=1e-6; int n;DB alp,h,x[maxn],R[maxn],fx1[maxn],fx2[maxn],fy1[maxn],fy2[maxn]; DB f(DB X){ DB s=0; for(int i=1;i<=n;++i){ if(fabs(x[i]-X)<R[i]) s=max(s,sqrt(sqr(R[i])-sqr(x[i]-X))); if(x[i+1]-x[i]>fabs(R[i+1]-R[i]) && fx1[i]<X && X<fx2[i]) s=max(s,(fy2[i]-fy1[i])/(fx2[i]-fx1[i])*(X-fx1[i])+fy1[i]); } return s; } DB simpson(DB a,DB b,DB fa,DB fb,DB fm){return (b-a)/6*(fa+4*fm+fb);}//不知道怎么积出来的辛普森 DB area(DB l,DB fl,DB m,DB fm,DB r,DB fr,DB pre){ DB ls=(l+m)/2,rs=(m+r)/2,fls=f(ls),frs=f(rs); DB la=simpson(l,m,fl,fm,fls),ra=simpson(m,r,fm,fr,frs); return fabs(la+ra-pre)<zero? pre:area(l,fl,ls,fls,m,fm,la)+area(m,fm,rs,frs,r,fr,ra); } int main(){ scanf("%d%lf",&n,&alp);alp=1/tan(alp); for(int i=1;i<n+2;++i) scanf("%lf",&x[i]),h+=x[i],x[i]=h*alp; DB l=x[n+1],r=l; for(int i=1;i<n+1;++i) scanf("%lf",&R[i]),l=l<x[i]-R[i]? l:x[i]-R[i],r=r>x[i]+R[i]? r:x[i]+R[i]; for(int i=1;i<n+1;++i) fx1[i]=x[i]+R[i]*(R[i]-R[i+1])/(x[i+1]-x[i]),fy1[i]=sqrt(sqr(R[i])-sqr(fx1[i]-x[i])), fx2[i]=x[i+1]+R[i+1]*(R[i]-R[i+1])/(x[i+1]-x[i]),fy2[i]=sqrt(sqr(R[i+1])-sqr(fx2[i]-x[i+1])); DB m=(l+r)/2,fm=f(m),fl=f(l),fr=f(r); printf("%.2lf",2*area(l,fl,m,fm,r,fr,simpson(l,r,fl,fr,fm))); return 0; }
E
//求图的绝对中心(这个点到所有点的最短距离的最大值最小) //reference：http://blog.csdn.net/crazy_ac/article/details/8816877 #include<cstdio> #include<iostream> using namespace std; const int N=205; const int inf=0x3f3f3f3f; int n,m,d[N][N],di[N][N],rk[N][N]; int main(){ scanf("%d%d",&n,&m); for(int i=1;i<=n;i++){ d[i][i]=di[i][i]=0; for(int j=i+1;j<=n;j++){ d[i][j]=d[j][i]=inf; di[i][j]=di[j][i]=inf; } } for(int i=1,x,y,z;i<=m;i++){ scanf("%d%d%d",&x,&y,&z); d[x][y]=d[y][x]=z; di[x][y]=di[y][x]=z; } for(int k=1;k<=n;k++){ for(int i=1;i<=n;i++){ for(int j=1;j<=n;j++){ d[i][j]=min(d[i][j],d[i][k]+d[k][j]); } } } for(int i=1;i<=n;i++){ for(int j=1;j<=n;j++) rk[i][j]=j; for(int j=1;j<=n;j++){ for(int k=j+1;k<=n;k++){ if(d[i][rk[i][j]]>d[i][rk[i][k]]){ swap(rk[i][j],rk[i][k]); } } } } int ans=inf; for(int i=1;i<=n;i++){ for(int j=1;j<=n;j++){ if(i==j) continue; ans=min(ans,d[i][rk[i][n]]<<1); ans=min(ans,d[j][rk[j][n]]<<1); for(int cmp=n,t=n-1;t>=1;t--){ if(d[j][rk[i][t]]>d[j][rk[i][cmp]]){ ans=min(ans,d[i][rk[i][t]]+d[j][rk[i][cmp]]+di[i][j]); cmp=t; } } } } printf("%.2lf\n",(double)ans/2); return 0; }
