Q

题意：求被三个或三个以上立方体重合的体积

分析：就是平面面积的加强，不过归根还是一样的，可以把z轴按照从小向大分区间N个，然后可以得到N个平面，用平面重复三次以上的在和高度计算体积。

************************************************************************

#include<stdio.h>
#include<algorithm>
using namespace std;

#define Lson r<<1
#define Rson r<<1|1

const int MAXN = 2005;

struct Edge{int x, y1, y2, dir;}e[MAXN];
struct point{int x1,x2, y1,y2, z1,z2;}p[MAXN];
struct segmentTree
{///cover 表示覆盖的次数
    int L, R, cover;
    long long len1, len2, len3;
    int Mid(){return (L+R)>>1;}
}a[MAXN<<2];
int Hash_Y[MAXN], ny, Hash_Z[MAXN], nz;

bool cmp(Edge n1, Edge n2)
{
    return n1.x < n2.x;
}
int  FindSegLen(int y1, int y2)
{
    return Hash_Y[y2] - Hash_Y[y1];
}
void BuildTree(int r, int L, int R)
{
    a[r].L = L, a[r].R = R;
    a[r].len1 = a[r].len2 = a[r].len3 = a[r].cover = 0;

    if(L == R-1)return ;

    BuildTree(Lson, L, a[r].Mid());
    BuildTree(Rson, a[r].Mid(), R);
}
void PushUp(int r)
{///合并，注意要最大区间更新
    if( a[r].cover > 2 )
        a[r].len1 = a[r].len2 = a[r].len3 = FindSegLen( a[r].L, a[r].R );
    else if( a[r].L == a[r].R-1 && a[r].cover == 2 )
        a[r].len3 = 0, a[r].len2 = a[r].len1 = FindSegLen( a[r].L, a[r].R );
    else if( a[r].L == a[r].R-1 && a[r].cover == 1 )
        a[r].len3 = a[r].len2 = 0, a[r].len1 = FindSegLen( a[r].L, a[r].R );
    else if( a[r].L == a[r].R-1 )
        a[r].len1 = a[r].len2 = a[r].len3 = 0;
    else if( a[r].cover == 2 )
    {
        a[r].len3 = a[Lson].len1 + a[Rson].len1;
        a[r].len2 = a[r].len1 = FindSegLen( a[r].L, a[r].R );;
    }
    else if( a[r].cover == 1 )
    {
        a[r].len3 = a[Lson].len2 + a[Rson].len2;
        a[r].len2 = a[Lson].len1 + a[Rson].len1;
        a[r].len1 = FindSegLen( a[r].L, a[r].R );
    }
    else
    {
        a[r].len3 = a[Lson].len3 + a[Rson].len3;
        a[r].len2 = a[Lson].len2 + a[Rson].len2;
        a[r].len1 = a[Lson].len1 + a[Rson].len1;
    }
}
void UpData(int r, int L, int R, int dir)
{
    if( a[r].L == L && a[r].R == R )
    {
        a[r].cover += dir;
        PushUp(r);

        return ;
    }

    if(R <= a[r].Mid())
        UpData(Lson, L, R, dir);
    else if(L >= a[r].Mid())
        UpData(Rson, L, R, dir);
    else
    {
        UpData(Lson, L, a[r].Mid(), dir);
        UpData(Rson, a[r].Mid(), R, dir);
    }

    PushUp(r);
}

int main()
{
    int T, t=1;

    scanf("%d", &T);

    while(T--)
    {
        int i, j, k, N;
        long long V=0;

        scanf("%d", &N);

        for(nz=i=0; i<N; i++)
        {
            scanf("%d%d%d%d%d%d", &p[i].x1,&p[i].y1,&p[i].z1,&p[i].x2,&p[i].y2,&p[i].z2);
            Hash_Z[nz++] = p[i].z1, Hash_Z[nz++] = p[i].z2;
        }

        sort(Hash_Z, Hash_Z+nz);
        nz = unique(Hash_Z, Hash_Z+nz) - Hash_Z;

        for(i=0; i<nz-1; i++)
        {///以z轴底部开始，判断每层是否有三个区间重合的
            for(ny=k=j=0; j<N; j++)
            {
                if( Hash_Z[i] >= p[j].z1 && Hash_Z[i] < p[j].z2 )
                {
                    e[k].x=p[j].x1, e[k].y1=p[j].y1, e[k].y2=p[j].y2, e[k++].dir=1;
                    e[k].x=p[j].x2, e[k].y1=p[j].y1, e[k].y2=p[j].y2, e[k++].dir=-1;
                    Hash_Y[ny++] = p[j].y1, Hash_Y[ny++] = p[j].y2;
                }
            }

            sort(Hash_Y, Hash_Y+ny);
            ny = unique(Hash_Y, Hash_Y+ny)-Hash_Y;
            BuildTree(1, 0, ny-1);

            sort(e, e+k, cmp);

            long long S = 0;

            for(j=0; j<k-1; j++)
            {
                int L = lower_bound(Hash_Y, Hash_Y+ny, e[j].y1) - Hash_Y;
                int R = lower_bound(Hash_Y, Hash_Y+ny, e[j].y2) - Hash_Y;

                UpData(1, L, R, e[j].dir);

                S += (long long)a[1].len3 * ( e[j+1].x - e[j].x );
            }

            V += S * (Hash_Z[i+1] - Hash_Z[i]);
        }

        printf("Case %d: %lld
", t++, V);
    }

    return 0;
}