hdu 3642 覆盖3次以上体积

http://www.cnblogs.com/kane0526/archive/2013/03/06/2947118.html

 

题目大意：给你n个立方体，求相交区域大于等于三次的体积和。

 

这题需要前面两题的知识

体积并

http://www.cnblogs.com/qlky/p/5759481.html

面积交：

http://www.cnblogs.com/qlky/p/5765617.html

 

对于面积交，另外建立len2数组表示覆盖2次的面积，len3数组表示覆盖3次及以上的面积

对于体积并，离散化x坐标是为了建树，离散化z坐标是为了节省时间。对于所有的体积范围进行面积交之和即为所求体积

 

#include <iostream>
#include <string>
#include <cstring>
#include <cstdlib>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <stack>
#include <queue>
#include <cctype>
#include <vector>
#include <iterator>
#include <set>
#include <map>
#include <sstream>
using namespace std;

#define mem(a,b) memset(a,b,sizeof(a))
#define pf printf
#define sf scanf
#define spf sprintf
#define pb push_back
#define debug printf("!\n")
#define MAXN 2000+5
#define MAX(a,b) a>b?a:b
#define blank pf("\n")
#define LL long long
#define ALL(x) x.begin(),x.end()
#define INS(x) inserter(x,x.begin())
#define pqueue priority_queue
#define INF 0x3f3f3f3f

#define ls (rt<<1)
#define rs (rt<<1|1)

int n,m;

int hh[MAXN],hh2[MAXN],col[MAXN<<3],len[MAXN<<3],len2[MAXN<<3],len3[MAXN<<3];

struct node
{
    int l,r,x,c;
    int z1,z2;
    node(){}
    node(int a,int b,int c,int d,int e,int f):l(a),r(b),x(c),c(d),z1(e),z2(f){}
    bool operator < (const node &b) const
    {
        return x<b.x;
    }
}a[MAXN<<3],tmp[MAXN<<3];

void PushUp(int rt,int l,int r)
{
    if(col[rt]>=3)
    {
        len[rt] = len2[rt] = len3[rt] = hh[r+1] - hh[l];
    }
    else if(col[rt] == 2)
    {
        len[rt] = len2[rt] = hh[r+1] - hh[l];
        if(l==r) len3[rt] = 0;
        else len3[rt] = len[ls]+len[rs];
    }
    else if(col[rt] == 1)
    {
        len[rt] = hh[r+1] - hh[l];
        if(l==r) len2[rt] = len3[rt] = 0;
        else
        {
            len2[rt] = len[ls]+len[rs];
            len3[rt] = len2[ls]+len2[rs];
        }
    }
    else if(l==r) len[rt] = len2[rt] = len3[rt] = 0;
    else
    {
        len[rt] = len[ls]+len[rs];
        len2[rt] = len2[ls]+len2[rs];
        len3[rt] = len3[ls]+len3[rs];
    }
}

void update(int val,int L,int R,int l,int r,int rt)
{
    if(L<=l && r<=R)
    {
        col[rt] += val;
        PushUp(rt,l,r);
        return;
    }
    int mid = (l+r)>>1;
    if(L <= mid) update(val,L,R,l,mid,ls);
    if(R > mid) update(val,L,R,mid+1,r,rs);
    PushUp(rt,l,r);
}

int main()
{
    int n,i,j,t,kase=1;
    sf("%d",&t);
    while(t--)
    {
        int v=0;
        sf("%d",&n);
        LL ans = 0;
        for(i=1;i<=n;i++)
        {
            int x1,x2,y1,y2;
            int z1,z2;
            sf("%d%d%d%d%d%d",&x1,&y1,&z1,&x2,&y2,&z2);
            hh[++v]=y1;
            hh2[v]=z1;
            a[v]=node(y1,y2,x1,1,z1,z2);
            hh[++v]=y2;
            hh2[v]=z2;
            a[v]=node(y1,y2,x2,-1,z1,z2);
        }
        sort(hh+1,hh+1+v);
        sort(hh2+1,hh2+1+v);
        sort(a+1,a+1+v);
        int d=1,d2=1;
        for(i=2;i<=v;i++)
            if(hh[i]!=hh[i-1])
                hh[++d]=hh[i];
        for(i=2;i<=v;i++)
            if(hh2[i]!=hh2[i-1])
                hh2[++d2]=hh2[i];
        int ct =0;
        for(j=1;j<=d2-1;j++)
        {
            ct=0;
            for(i=1;i<=v;i++)
                if(a[i].z1<=hh2[j] && hh2[j]< a[i].z2)
                    tmp[ct++]=a[i];
            mem(col,0);
            mem(len,0);
            mem(len2,0);
            mem(len3,0);
            for(i=0;i<ct-1;i++)
            {
                int l = lower_bound(hh+1,hh+d,tmp[i].l)-hh;
                int r = lower_bound(hh+1,hh+d,tmp[i].r)-hh-1;
                if(l<=r) update(tmp[i].c,l,r,1,d,1);
                //pf("t%d %d %d %d %d\n",len3[1],hh2[j+1],hh2[j],tmp[i+1].x,tmp[i].x);
                ans+=(LL)len3[1]*(LL)(hh2[j+1]-hh2[j])*(tmp[i+1].x-tmp[i].x);
                //pf("t%d\n",len3[1]);
            }
        }
        pf("Case %d: %I64d\n",kase++,ans);
    }
    return 0;
}