POJ1151-扫面线+线段树+离散化//入门题

比较水的入门题

记录矩形竖边的x坐标，离散化排序。以被标记的边建树。

扫描线段树，查询线段树内被标记的边。遇到矩形的右边就删除此边

每一段的面积是查询结果乘边的横坐标之差，求和就是答案

 #include <cstdio>
 #include <cstring>
 #include <algorithm>

 using namespace std;
 const int maxn = ;
 int N,num,kase;
 double savey[maxn*];

 struct line{
     double x,y1,y2;
     int flag;
     bool operator < (const struct line &t) const {return x < t.x;}
 }Line[maxn];

 struct Node{
     int l,r;
     double dl,dr;
     double len;
     int flag;
 }SegTree[maxn*];

 void Build(int i,int l,int r)
 {
     SegTree[i].l = l;
     SegTree[i].r = r;
     SegTree[i].flag = SegTree[i].len = ;
     SegTree[i].dl = savey[l];
     SegTree[i].dr = savey[r];
     if(l + == r) return;
     int mid = (l+r)>>;
     Build(i<<,l,mid);
     Build(i<<|,mid,r);
 }

 void getlen(int t)
 {
     if(SegTree[t].flag > )
     {
         SegTree[t].len = SegTree[t].dr - SegTree[t].dl;
         return ;
     }
     if(SegTree[t].l+ == SegTree[t].r) SegTree[t].len = ;
     else SegTree[t].len = SegTree[t<<].len + SegTree[t<<|].len;
 }

 void Update(int i,line e)
 {
     if(e.y1 == SegTree[i].dl && e.y2 == SegTree[i].dr)
     {
         SegTree[i].flag += e.flag;
         getlen(i);
         return;
     }
     if(e.y2 <= SegTree[i<<].dr) Update(i<<,e);
     else if(e.y1 >= SegTree[i<<|].dl)    Update(i<<|,e);
     else
     {
         line temp = e;
         temp.y2 = SegTree[i<<].dr;
         Update(i<<,temp);
         temp = e;
         temp.y1 = SegTree[i<<|].dl;
         Update(i<<|,temp);
     }
     getlen(i);
 }

 int main()
 {
     while(~scanf("%d",&N) && N)
     {
         kase++;
         num=;
         double x1,x2,y1,y2;
         for(int i=;i<=N;i++)
         {
             scanf("%lf%lf%lf%lf",&x1,&y1,&x2,&y2);
             Line[num].x = x1;
             Line[num].y1 = y1;
             Line[num].y2 = y2;
             Line[num].flag = ;
             savey[num++]=y1;
             Line[num].x = x2;
             Line[num].y1 = y1;
             Line[num].y2 = y2;
             Line[num].flag = -;
             savey[num++] = y2;
         }
         sort(Line+,Line+num);
         sort(savey+,savey+num);
         Build(,,num-);
         Update(,Line[]);
         double ans = ;
         for(int i=;i<num;i++)
         {
             //printf("%f %f\n",SegTree[1].len,Line[i].x-Line[i-1].x);
             ans += SegTree[].len * (Line[i].x - Line[i-].x);
             Update(,Line[i]);
         }
         printf("Test case #%d\n",kase);
         printf("Total explored area: %.2f\n\n",ans);
     }
     return ;
 }