poj1151Atlantis(离散化+扫描线）

http://poj.org/problem?id=1151

http://www.cnblogs.com/kane0526/archive/2013/02/26/2934214.html这篇博客写的不错 主要是图画的不错

求面积并 离散化后 通过添加矩形的x方向边  用线段树不断更新（要求的分割开的）矩形的长和宽

 #include <iostream>
 #include<cstdio>
 #include<algorithm>
 #include<cstring>
 #define maxn 5050
 using namespace std;
 struct node
 {
     double lx,rx,y;
     int s;
     node(){}
     node(double a,double b,double c,int d):lx(a),rx(b),y(c),s(d){}
     bool operator < (const node &S) const
     {
         return y<S.y;
     }
 }mat[maxn];
 double que[maxn],sum[maxn];
 int cnt[maxn];
 int bin(double x,int n)
 {
     int s,e,m;
     s=;e=n;
     while(s<=e)
     {
         m = (s+e)/;
         if(que[m]==x)
         return m;
         else if(que[m]>x)
         e = m-;
         else
         s = m+;
     }
     return m;
 }
 void pushup(int l,int r,int w)
 {
     if(cnt[w])
     sum[w] = que[r+]-que[l];
     else if(l==r)
     sum[w] = ;
     else
     sum[w] = sum[w<<]+sum[w<<|];
 }
 void update(int a,int b,int d,int l,int r,int w)
 {
     if(a<=l&&b>=r)
     {
         cnt[w]+=d;
         pushup(l,r,w);
         return ;
     }
     int m = (l+r)>>;
     if(a<=m)
     update(a,b,d,l,m,w<<);
     if(b>m)
     update(a,b,d,m+,r,w<<|);
     pushup(l,r,w);
 }
 int main()
 {
     int i,k,n,cas=;
     double a,b,c,d;
     while(cin>>n)
     {
         if(n==) break;
         int num = ;
         for(i = ; i <= n ; i++)
         {
             cin>>a>>b>>c>>d;
             que[num] = a;
             mat[num++]=node(a,c,b,);
             que[num] = c;
             mat[num++] = node(a,c,d,-);
         }
         sort(que,que+num);
         sort(mat,mat+num);
         k = ;
         for(i =  ; i < num ; i++)
         {
             if(que[i]!=que[i-])
             que[k++] = que[i];
         }
         double ans=;
         memset(cnt,,sizeof(cnt));
         memset(sum,,sizeof(sum));
         for(i =  ; i < num- ; i++)
         {
             int l = bin(mat[i].lx,k-);
             int r = bin(mat[i].rx,k-)-;
             if(l<=r)
             update(l,r,mat[i].s,,k-,);
             ans+=sum[]*(mat[i+].y-mat[i].y);
         }
         printf("Test case #%d\nTotal explored area: %.2f\n\n",cas++,ans);
     }
     return ;
 }