hdu 3642 Get The Treasury
Get The Treasury
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1036 Accepted Submission(s): 318
Jack can’t get the total volume of the treasury because these regions don’t always contain treasury. Through years of experience, he discovers that if a region is detected that may have treasury at more than two different spots, the region really exist treasure. And now Jack only wants to know the minimum volume of the treasury.
Now Jack entrusts the problem to you.
Each test case is given in some lines. In the first line there is an integer n (1 ≤ n ≤ 1000), the number of spots on the surface of the earth that he had detected. Then n lines follow, every line contains six integers x1, y1, z1, x2, y2 and z2, separated by a space. The absolute value of x and y coordinates of the vertices is no more than 106, and that of z coordinate is no more than 500.
//求矩形交3次以上部分体积
//枚举Z轴然后线段树算面积
//搞了一下午、加深了对求面积题目的理解呀
#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <queue>
#include <stack>
#define N 2003
#define lson l,m,k<<1
#define rson m,r,k<<1|1
using namespace std;
struct node
{
int x,y1,y2,z1,z2;
int flag;
bool operator<(const node&a)const
{
return x<a.x;
}
};
struct tree
{
int cover,one,two,more;//代表的是当前区间被覆盖1,2,2次以上线段长度
};
int rcy[N],rcz[N];
node In[N];
tree st[N<<2];
void build(int l,int r,int k)
{
st[k].cover=st[k].one=st[k].two=st[k].more=0;
if(l+1==r)
return;
int m=(l+r)>>1;
build(lson);
build(rson);
}
void up(int &k,int &l,int &r)//这个是这类题目的关键
{
int ls=k<<1,rs=k<<1|1;
if(st[k].cover>2)
{
st[k].more=rcy[r]-rcy[l];
st[k].one=st[k].two=0;
return;
}
if(st[k].cover==2)
{
if(l+1==r){st[k].two=rcy[r]-rcy[l];st[k].one=st[k].more=0;return ;}
st[k].more=st[ls].one+st[rs].one+st[ls].two+st[rs].two;
st[k].more+=st[ls].more+st[rs].more;
st[k].two=rcy[r]-rcy[l]-st[k].more;
st[k].one=0;
return ;
}
if(st[k].cover==1)
{
if(l+1==r){st[k].one=rcy[r]-rcy[l];st[k].two=st[k].more=0;return ;}
st[k].two=st[ls].one+st[rs].one;
st[k].more=st[ls].two+st[rs].two+st[ls].more+st[rs].more;
st[k].one=rcy[r]-rcy[l]-st[k].more-st[k].two;
return ;
}
if(l+1==r){ st[k].one=st[k].two=st[k].more=0;return;}
st[k].one=st[ls].one+st[rs].one;
st[k].two=st[ls].two+st[rs].two;
st[k].more=st[ls].more+st[rs].more;
}
int flag;
void update(int &y1,int &y2,int l,int r,int k)
{
if(y1<=rcy[l]&&y2>=rcy[r])
{
st[k].cover+=flag;
up(k,l,r);
return;
}
int m=(l+r)>>1;
if(y1<rcy[m]) update(y1,y2,lson);
if(y2>rcy[m]) update(y1,y2,rson);
up(k,l,r);
}
int main()
{
int x1,y1,z1,x2,y2,z2;
int n,t=1,T;
int i,j,k,l,ry,rz;
double v,s;
scanf("%d",&T);
while(T--)
{
scanf("%d",&n);
for(j=i=0;i<n;i++)
{
scanf("%d%d%d%d%d%d",&x1,&y1,&z1,&x2,&y2,&z2);
In[j].x=x1;In[j].y1=y1;In[j].y2=y2;
In[j].z1=z1;In[j].z2=z2;
rcy[j]=y1;rcz[j]=z1; In[j++].flag=1;
In[j].x=x2;In[j].y1=y1;In[j].y2=y2;
In[j].z1=z1;In[j].z2=z2;
rcy[j]=y2;rcz[j]=z2; In[j++].flag=-1;
}
sort(In,In+j);
sort(rcy,rcy+j);
sort(rcz,rcz+j);
for(ry=0,i=1;i<j;i++)
if(rcy[i]!=rcy[ry])
rcy[++ry]=rcy[i];
for(rz=0,i=1;i<j;i++)
if(rcz[i]!=rcz[rz])
rcz[++rz]=rcz[i];
v=0;j--;
for(i=0;i<rz;i++)
{
build(0,ry,1);
s=0;
for(k=0;k<j;k++)
{
if(In[k].z1<=rcz[i]&&In[k].z2>rcz[i])
{
flag=In[k].flag;
update(In[k].y1,In[k].y2,0,ry,1);
for(l=k+1;l<j+1;l++)//开始没发现这个问题,Wa的我好郁闷
if(In[l].z1<=rcz[i]&&In[l].z2>rcz[i])
break;
s+=1.0*st[1].more*(In[l].x-In[k].x);
}
}
v+=s*(rcz[i+1]-rcz[i]);
}
printf("Case %d: %.0lf\n",t++,v);
}
return 0;
}