Get The Treasury

本文链接：https://blog.youkuaiyun.com/u014733623/article/details/41805599

本文介绍了一种解决三维空间中多个长方体区域重叠问题的方法，通过离散化和转换为二维问题来计算至少三次重叠的体积总和。采用线段树和扫描线算法实现，并给出具体代码实例。

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Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2083 Accepted Submission(s): 640

Problem Description

Jack knows that there is a great underground treasury in a secret region. And he has a special device that can be used to detect treasury under the surface of the earth. One day he got outside with the device to ascertain the treasury. He chose many different locations on the surface of the earth near the secret region. And at each spot he used the device to detect treasury and got some data from it representing a region, which may contain treasury below the surface. The data from the device at each spot is six integers x₁, y₁, z₁, x₂, y₂ and z₂ (x₁<x₂, y₁<y₂, z₁<z₂). According to the instruction of the device they represent the range of x, y and z coordinates of the region. That is to say, the x coordinate of the region, which may contain treasury, ranges from x₁ to x₂. So do y and z coordinates. The origin of the coordinates is a fixed point under the ground.
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.

Input

The first line of the input file contains a single integer t, the number of test cases, followed by the input data for each test case.
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.

Output

For each test case, you should output “Case a: b” in a single line. a is the case number, and b is the minimum volume of treasury. The case number is counted from one.

Sample Input

2
1
0 0 0 5 6 4
3
0 0 0 5 5 5
3 3 3 9 10 11
3 3 3 13 20 45

Sample Output

Case 1: 0
Case 2: 8

题意：给你n个长方体，问重复三次以上的面积是多少。

思路：因为z轴不超过500，所以要离散化z轴以后，转化成二维的来做。这样就和HDU1255差不多了。

AC代码如下：

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
typedef long long ll;
struct Cube
{
    ll x1,y1,z1,x2,y2,z2;
}cubes[10100];
struct node
{
    int l,r,lazy;
    ll sum[4];
}tree[16010];
struct node2
{
    int l,r,f;
    ll h;
}line[40100];
bool cmp(node2 a,node2 b)
{
    return a.h<b.h;
}
int n,mx,mz;
ll pz[40100],px[40100],fx[2000010];
void build(int o,int l,int r)
{
    tree[o].l=l;
    tree[o].r=r;
    tree[o].lazy=0;
    for(int i=1;i<=3;i++)
       tree[o].sum[i]=0;
    tree[o].sum[0]=px[tree[o].r]-px[tree[o].l-1];
    if(l==r)
      return;
    int mi=(l+r)/2;
    build(o<<1,l,mi);
    build(o<<1|1,mi+1,r);
}
void solve(int o)
{
    int i,j,k;
    if(tree[o].l==tree[o].r)
    {
        for(i=1;i<=min(tree[o].lazy,3);i++)
           tree[o].sum[i]=tree[o].sum[0];
        for(;i<=3;i++)
           tree[o].sum[i]=0;
        return;
    }
    if(tree[o].lazy>=3)
    {
        for(i=1;i<=3;i++)
           tree[o].sum[i]=tree[o].sum[0];
    }
    else if(tree[o].lazy==2)
    {
        for(i=1;i<=2;i++)
           tree[o].sum[i]=tree[o].sum[0];
        tree[o].sum[3]=tree[o<<1].sum[1]+tree[o<<1|1].sum[1];
    }
    else if(tree[o].lazy==1)
    {
        tree[o].sum[1]=tree[o].sum[0];
        tree[o].sum[2]=tree[o<<1].sum[1]+tree[o<<1|1].sum[1];
        tree[o].sum[3]=tree[o<<1].sum[2]+tree[o<<1|1].sum[2];
    }
    else if(tree[o].lazy==0)
    {
        for(i=1;i<=3;i++)
           tree[o].sum[i]=tree[o<<1].sum[i]+tree[o<<1|1].sum[i];
    }
}
void update(int o,int l,int r,int f)
{
    if(tree[o].l==l && tree[o].r==r && tree[o].lazy+f>=0)
    {
        tree[o].lazy+=f;
        solve(o);
        return;
    }
    int mi=(tree[o].l+tree[o].r)/2;
    if(r<=mi)
      update(o<<1,l,r,f);
    else if(l>mi)
      update(o<<1|1,l,r,f);
    else
    {
        update(o<<1,l,mi,f);
        update(o<<1|1,mi+1,r,f);
    }
    solve(o);
}
int main()
{
    int T,t,i,j,k,p,z,l,r;
    ll ans;
    scanf("%d",&T);
    for(t=1;t<=T;t++)
    {
        scanf("%d",&n);
        for(i=1;i<=n;i++)
        {
            scanf("%I64d%I64d%I64d%I64d%I64d%I64d",&cubes[i].x1,&cubes[i].y1,&cubes[i].z1,
                                                   &cubes[i].x2,&cubes[i].y2,&cubes[i].z2);
            cubes[i].x1+=1000000;cubes[i].x2+=1000000;
            pz[i*2-1]=cubes[i].z1;pz[i*2]=cubes[i].z2;
            px[i*2-1]=cubes[i].x1;px[i*2]=cubes[i].x2;
        }
        sort(pz+1,pz+1+n*2);
        sort(px+1,px+1+n*2);
        mx=-1;mz=-1;px[0]=-100000000;pz[0]=-100000000;
        for(i=1;i<=n*2;i++)
        {
            if(px[i]!=px[i-1])
            {
                px[++mx]=px[i];
                fx[px[i]]=mx;
            }
            if(pz[i]!=pz[i-1])
              pz[++mz]=pz[i];
        }
        ans=0;
        for(i=1;i<=mz;i++)
        {

            p=0;
            for(j=1;j<=n;j++)
               if(cubes[j].z1<pz[i] && cubes[j].z2>=pz[i])
               {
                   p+=2;
                   l=fx[cubes[j].x1]+1;r=fx[cubes[j].x2];
                   line[p-1].l=l;line[p-1].r=r;line[p-1].f=1;line[p-1].h=cubes[j].y1;
                   line[p].l=l;line[p].r=r;line[p].f=-1;line[p].h=cubes[j].y2;
               }
            sort(line+1,line+1+p,cmp);
            build(1,1,mx);
            for(j=1;j<p;j++)
            {
                update(1,line[j].l,line[j].r,line[j].f);
                ans+=(pz[i]-pz[i-1])*tree[1].sum[3]*(line[j+1].h-line[j].h);
            }
        }
        printf("Case %d: %I64d\n",t,ans);
    }
}