hdu 3642 Get The Treasure（掃描線）

題意：有T給測試資料，每組資料先給一個數字N，接下來的N行裡，每行裡有6個數字，分別是x1,y1,z1,x2,y2,z2，表示這個長方體x軸方向的範圍從x1到x2，y座標和z座標類似，求至少有三個長方體相交的體積是多少。

        因為Z軸的範圍很小，所以我們將Z軸離散化之後，枚舉Z[i]和Z[i+1]之間，矩形並時覆蓋了三次以上的面積，那麼這時候，就可以求出在Z[i]和Z[i+1]之間題目所求的體積，遍曆一次Z座標，也就得出了答案。如果求矩形並時覆蓋超過三次的面積，與hdu 1255 覆蓋的面積，是很類似的。

#include <iostream>#include <cstdio>#include <cstring>#include <vector>#include <algorithm>#include <map>using namespace std;#define LL(x) (x<<1)#define RR(x) (x<<1|1)#define MID(a,b) (a+((b-a)>>1))typedef long long LL;const int N=2005;struct CUBE{    int x1,y1,z1;    int x2,y2,z2;    void get()    {        scanf("%d%d%d",&x1,&y1,&z1);        scanf("%d%d%d",&x2,&y2,&z2);    }}cube[N];struct Line{    int x,y1,y2,flag;    Line(){}    Line(int a,int b,int c,int d)    { x=a;y1=b;y2=c;flag=d; }    bool operator<(const Line &b)const    { return x<b.x; }};struct node{    int lft,rht,flag,len[4];    int mid(){return MID(lft,rht);}    void init(){memset(len,0,sizeof(len));}};vector<int> y,z;vector<Line> line;map<int,int> H;struct Segtree{    node tree[N*4];    void calu(int ind)    {        if(tree[ind].flag>=3)        {            tree[ind].len[3]=tree[ind].len[0];            tree[ind].len[2]=tree[ind].len[1]=0;        }        else if(tree[ind].flag==2)        {            tree[ind].len[2]=tree[ind].len[0];            if(tree[ind].lft+1==tree[ind].rht)            {                tree[ind].len[1]=tree[ind].len[3]=0;            }            else            {                tree[ind].len[3]=tree[LL(ind)].len[3]+tree[RR(ind)].len[3]                    +tree[LL(ind)].len[2]+tree[RR(ind)].len[2]                    +tree[LL(ind)].len[1]+tree[RR(ind)].len[1];                tree[ind].len[1]=0;                tree[ind].len[2]-=tree[ind].len[3];            }        }        else if(tree[ind].flag==1)        {            tree[ind].len[1]=tree[ind].len[0];            if(tree[ind].lft+1==tree[ind].rht)            {                tree[ind].len[2]=tree[ind].len[3]=0;            }            else            {                tree[ind].len[3]=tree[LL(ind)].len[3]+tree[RR(ind)].len[3]                    +tree[LL(ind)].len[2]+tree[RR(ind)].len[2];                tree[ind].len[2]=tree[LL(ind)].len[1]+tree[RR(ind)].len[1];                tree[ind].len[1]-=(tree[ind].len[2]+tree[ind].len[3]);            }        }        else        {            if(tree[ind].lft+1==tree[ind].rht)            {                tree[ind].len[1]=tree[ind].len[2]=tree[ind].len[3]=0;            }            else            {                tree[ind].len[3]=tree[LL(ind)].len[3]+tree[RR(ind)].len[3];                tree[ind].len[2]=tree[LL(ind)].len[2]+tree[RR(ind)].len[2];                tree[ind].len[1]=tree[LL(ind)].len[1]+tree[RR(ind)].len[1];            }        }    }    void build(int lft,int rht,int ind)    {        tree[ind].lft=lft;    tree[ind].rht=rht;        tree[ind].init();    tree[ind].flag=0;        tree[ind].len[0]=y[rht]-y[lft];        if(lft+1!=rht)        {            int mid=tree[ind].mid();            build(lft,mid,LL(ind));            build(mid,rht,RR(ind));        }    }    void updata(int st,int ed,int ind,int valu)    {        int lft=tree[ind].lft,rht=tree[ind].rht;        if(st<=lft&&rht<=ed) tree[ind].flag+=valu;        else        {            int mid=tree[ind].mid();            if(st<mid) updata(st,ed,LL(ind),valu);            if(ed>mid) updata(st,ed,RR(ind),valu);        }        calu(ind);    }}seg;int main(){    int t,t_cnt=0;    scanf("%d",&t);    while(t--)    {        y.clear(); z.clear(); line.clear(); H.clear();        int n;        scanf("%d",&n);        for(int i=0;i<n;i++)        {            cube[i].get();            y.push_back(cube[i].y1); y.push_back(cube[i].y2);            z.push_back(cube[i].z1); z.push_back(cube[i].z2);        }        printf("Case %d: ",++t_cnt);        if(n<3) {puts("0");continue;}        else        {            sort(y.begin(),y.end());            sort(z.begin(),z.end());            y.erase(unique(y.begin(),y.end()),y.end());            z.erase(unique(z.begin(),z.end()),z.end());            for(int i=0;i<(int)y.size();i++) H[y[i]]=i;            LL res=0;            seg.build(0,(int)y.size()-1,1);            for(int i=0;i<(int)z.size()-1;i++)            {                line.clear();                for(int j=0;j<n;j++)                {                    if(cube[j].z1<=z[i]&&cube[j].z2>=z[i+1])                    {                        line.push_back(Line(cube[j].x1,cube[j].y1,cube[j].y2,1));                        line.push_back(Line(cube[j].x2,cube[j].y1,cube[j].y2,-1));                    }                }                sort(line.begin(),line.end());                for(int j=0;j<(int)line.size();j++)                {                    if(j!=0) res+=(z[i+1]-z[i])*(line[j].x-line[j-1].x)*(LL)seg.tree[1].len[3];                    seg.updata(H[line[j].y1],H[line[j].y2],1,line[j].flag);                }            }            printf("%I64d\n",res);        }    }    return 0;}