sgu Kalevich Strikes Back

这道题就是求一个大矩形被n个矩形划分成n+1个部分的面积，这些矩形之间不会相交，可能包含。。

 #include <cstdio>
 #include <cstring>
 #include <vector>
 #include <algorithm>
 #define maxn 120100
 using namespace std;

 long long s[maxn];
 vector<int>g[maxn];
 int pre[maxn];
 int X[maxn];
 int n,m,w,h,x1,y1,x2,y2;
 struct node
 {
     int l,r;
     int ll,rr;
     int cover;
 }tree[maxn*];

 struct line
 {
     int y,x1,x2,lr;
     bool operator <(const line &a)const
     {
         return y<a.y;
     }
 }p[maxn];

 void build(int i,int l,int r)
 {
     tree[i].l=l;
     tree[i].r=r;
     tree[i].ll=X[l];
     tree[i].rr=X[r];
     tree[i].cover=;
     if(r-l==) return ;
     int mid=(l+r)>>;
     build(i<<,l,mid);
     build(i<<|,mid,r);
 }
 void down(int i)
 {
     if(tree[i].l+==tree[i].r) return;
     if(tree[i].cover!=-)
     {
         tree[i<<].cover=tree[i<<|].cover=tree[i].cover;
         tree[i].cover=-;
     }
 }
 void update(int i, line a)
 {
     if(tree[i].ll==a.x1&&tree[i].rr==a.x2)
     {
         if(a.lr>)
             tree[i].cover=a.lr;
         else
             tree[i].cover=pre[-a.lr];
         return ;
     }
     down(i);
     if(a.x2<=tree[i<<].rr) update(i<<,a);
     else if(a.x1>=tree[i<<|].ll) update(i<<|,a);
     else
     {
         line tmp=a;
         tmp.x2=tree[i<<].rr;
         update(i<<,tmp);
         tmp=a;
         tmp.x1=tree[i<<|].ll;
         update(i<<|,tmp);
     }
 }

 int search1(int i,line m)
 {
     if(tree[i].cover!=-)
     {
         return tree[i].cover;
     }
     if(m.x2<=tree[i<<].rr) return search1(i<<,m);
     else if(m.x1>=tree[i<<|].ll) return search1(i<<|,m);
     else
     {
         line tmp;
         tmp=m;
         tmp.x2=tree[i<<].rr;
         return search1(i<<,tmp);
     }
 }
 void dfs(int u)
 {
     for(int i=; i<(int)g[u].size(); i++)
     {
         int v=g[u][i];
         s[u]-=s[v];
         dfs(v);
     }
 }

 int main()
 {
     while(scanf("%d",&n)!=EOF)
     {
         for(int i=; i<=n; i++)
         {
             g[i].clear();
         }
         scanf("%d%d",&w,&h);
         s[]=(long long)w*h;
         int t1=;
         for(int i=; i<=n; i++)
         {
             scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
             if(x1>x2)swap(x1,x2);
             if(y1>y2)swap(y1,y2);
             s[i]=(long long)(x2-x1)*(y2-y1);
             p[t1].y=y1;
             p[t1].x1=x1;
             p[t1].x2=x2;
             p[t1].lr=i;
             X[t1++]=x1;
             p[t1].y=y2;
             p[t1].x1=x1;
             p[t1].x2=x2;
             p[t1].lr=-i;
             X[t1++]=x2;
         }
         sort(X,X+t1);
         t1=unique(X,X+t1)-X;
         build(,,t1-);
         sort(p,p+*n);
         for(int i=; i<*n; i++)
         {
             if(p[i].lr>)
             {
                 pre[p[i].lr]=search1(,p[i]);
                 g[pre[p[i].lr]].push_back(p[i].lr);
             }
             update(,p[i]);
         }
         dfs();
         sort(s,s+n+);
         for(int i=; i<=n; i++)
         {
             if(i==n) printf("%I64d\n",s[i]);
             else printf("%I64d ",s[i]);
         }
     }
     return ;
 }