ZOJ problem Set-3664 Split the Rectangle

The ACM-ICPC Asia Changchun regional contest-j

The title of a freshman is to remove some of the two points in the same rectangle, and then ask how many rectangles are left.

Actually is can make a binary tree, the process of formation is the process of separating the rectangle. Then the answer is clearly the number of leaf nodes---A and B LCA is the root of the subtree of the number of leaf nodes +1.

#include <iostream> #include <cstdio> #include <cstdlib> #include <cstring> #define MAXN usi
NG namespace Std;
        struct rect{int x1,x2,y1,y2;
int l,r,soncnt;
}R[MAXN]; struct segment{int x1,x2,y1,y2;}
P[MAXN]; struct point{int x, y;}
b;
int N,q,tot; BOOL Innerline (int v,int l,int R) {return (v>=l && v<=r);} int Buildtree (Rect node, int fl
        AG) {R[tot]=node;
        R[tot].l=r[tot].r=-1;
        r[tot].soncnt=0;
        int res = tot++; for (int i=0;i<n;++i) if ((p[i].y1==p[i].y2) ^flag) {if (!flag) {if (p[i].x1==node.x 1 && p[i].x2==node.x2 && innerline (p[i].y1,node.y1,node.y2)) {Rect Te
                    MP = node;
                    Temp.y1=p[i].y1;
                    
                    R[res].l= Buildtree (Temp,!flag);
                    temp = node;
            Temp.y2=p[i].y2;        R[res].r= Buildtree (Temp,!flag); }} else {if (p[i].y1==node.y1 && p[i].y2==node.y2 && innerline (p[
                    I].X1,NODE.X1,NODE.X2)) {Rect temp = node;
                    temp.x1=p[i].x1;
                    
                    R[res].l= Buildtree (Temp,!flag);
                    temp= node;
                    temp.x2=p[i].x2;
                 R[res].r= Buildtree (Temp,!flag);
            }}} if (R[res].l==-1 && r[res].r==-1) R[res].soncnt=1;else {
            if (r[res].l!=-1) r[res].soncnt=r[r[res].l].soncnt;
        if (r[res].r!=-1) r[res].soncnt+=r[r[res].r].soncnt;
} return res; } bool Innerrect (point C,rect R) {return (c.x>=r.x1 && c.x<=r.x2 && c.y>=r.y1 &&am P
C.Y&LT;=R.Y2); } int DFS (Rect node) {if (node.l!=-1) {if (Innerrect (A, R[NODE.L]) && innerrect (B,R[NODE.L])) return DFS (R[NODE.L]);  } if (Node.r!=-1) {if (Innerrect (A,R[NODE.R]) && innerrect (B,R[NODE.R])) return
        DFS (R[NODE.R]);
} return node.soncnt; } int main () {while (cin>>r[0].x1>>r[0].y1>>r[0].x2>>r[0].y2) {cin>>n>>
          Q;
              for (int i=0;i<n;++i) {scanf ("%d%d%d%d", &p[i].x1,&p[i].y1,&p[i].x2,&p[i].y2);
              if (p[i].x1>p[i].x2) swap (P[I].X1,P[I].X2);
          if (p[i].y1>p[i].y2) swap (P[I].Y1,P[I].Y2);
          } tot=0;
          Buildtree (r[0],0);
              for (int i=0;i<q;++i) {scanf ("%d%d%d%d", &a.x,&a.y,&b.x,&b.y);
          Cout<<n+2-dfs (r[0]) <<endl;
}} return 0; }