HDU 1255 覆盖的面积 (线段树+扫描线+离散化)

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1255

题意很清楚，就是让你求矩阵之间叠加层数大于1的矩形块的面积和。

因为n只有1000，所以我离散化一下，数据大小就缩小了，那么之后只需要线段树单点更新就好了。

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <map>
 #include <algorithm>
 using namespace std;
 const int MAXN = 4e3 + ;
 struct data {
     int x1 , x2 , flag;
     double y , xx1 , xx2;
     bool operator <(const data& cmp) const {
         return y < cmp.y;
     }
 }a[MAXN];
 struct segtree {
     int l , r , add;
     double val;
 }T[MAXN << ];
 double x[MAXN];
 map <double , int> mp;

 void build(int p , int l , int r) {
     int mid = (l + r) >> ;
     T[p].val = T[p].add =  , T[p].l = l , T[p].r = r;
     if(r - l == ) {
         return ;
     }
     build(p <<  , l , mid);
     build((p << )| , mid , r);
 }

 void updata(int p , int l , int r , int val) {
     int mid = (T[p].l + T[p].r) >> ;
     if(T[p].r - T[p].l == ) {
         T[p].add += val;
         if(T[p].add > )
             T[p].val = x[T[p].r] - x[T[p].l];
         else
             T[p].val = ;
         return ;
     }
     if(r <= mid) {
         updata(p <<  , l , r , val);
     }
     else if(l >= mid) {
         updata((p << )| , l , r , val);
     }
     else {
         updata(p <<  , l , mid , val);
         updata((p << )| , mid , r , val);
     }
     T[p].val = T[p << ].val + T[(p << )|].val;
 }

 int main()
 {
     int t =  , n;
     double x1 , x2 , y1 , y2;
     scanf("%d" , &t);
     while(t--) {
         mp.clear();
         scanf("%d" , &n);
         int cnt = ;
         for(int i =  ; i < n ; i++) {
             scanf("%lf %lf %lf %lf" , &x1 , &y1 , &x2 , &y2);
             if(x1 > x2)
                 swap(x1 , x2);
             if(y1 > y2)
                 swap(y1 , y2);
             int ls = i *  , rs = i *  + ;
             a[ls].xx1 = x1 , a[ls].xx2 = x2 , a[ls].y = y1 , a[ls].flag = ;
             a[rs].xx1 = x1 , a[rs].xx2 = x2 , a[rs].y = y2 , a[rs].flag = -;
             if(!mp[x1]) {
                 mp[x1] = ;
                 x[++cnt] = x1;
             }
             if(!mp[x2]) {
                 mp[x2] = ;
                 x[++cnt] = x2;
             }
         }
         sort(a , a + n * );
         sort(x +  , x + cnt + );
         for(int i =  ; i < n *  ; i++) {
             a[i].x1 = lower_bound(x +  , x + cnt +  , a[i].xx1) - x;
             a[i].x2 = lower_bound(x +  , x + cnt +  , a[i].xx2) - x;
         }
         double res = ;
         build( ,  , cnt);
         updata( , a[].x1 , a[].x2 , a[].flag);
         for(int i =  ; i <  * n ; i++) {
             res += (a[i].y - a[i - ].y) * T[].val;
             updata( , a[i].x1 , a[i].x2 , a[i].flag);
         }
         printf("%.2f\n" , res);
     }
 }