hdu1255 扫描线，矩形重叠面积（两次以上）

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题意：
给你n个矩形，然后问你这n个矩形所组成的画面中被覆盖至少两次的面积有多大。
思路：
和1542差距并不是很大，大体上还是离散化+线段树扫面线，不同的地方就是这个题目要求覆盖至少两次，那么假如l1:覆盖一次的区间长度,l2：覆盖至少两次的区间长度， l3：整个区间的长度，并且满足 l1 + l2 = l3,cnt为区间覆盖次数，那么在更新pushup的时候
(1)cnt >= 2 那么l2 = l3 ,l1 = 0
(2)cnt == 1 那么l2 = 左边l1 + 右边l1 + 左边l2 + 右边l2,l1 = l3 - l2

(3)cnt == 0 那么如果是叶子了l1 = l2 = 0,否则正常更新l1=l1左+l1右，l2=l2左+l2右.

#include<stdio.h>#include<string.h>#include<algorithm>#define lson l ,mid ,t << 1#define rson mid ,r ,t << 1 | 1#define N_node 10000using namespace std;typedef struct{   double l ,r ,h;   int mk;}EDGE;EDGE edge[N_node];double len1[N_node] ,len2[N_node];int cnt[N_node];double tmp[N_node] ,num[N_node];bool camp(EDGE a ,EDGE b){   return a.h < b.h;}void Pushup(int l ,int r ,int t){   if(cnt[t] >= 2)   {      len2[t] = num[r] - num[l];      len1[t] = 0;   }   else if(cnt[t] == 1)   {      len2[t] = len2[t<<1] + len2[t<<1|1] + len1[t<<1] + len1[t<<1|1];      len1[t] = num[r] - num[l] - len2[t];   }   else    {      if(l + 1 == r)      {         len2[t] = len1[t] = 0;      }      else       {         len2[t] = len2[t<<1] + len2[t<<1|1];         len1[t] = len1[t<<1] + len1[t<<1|1];      }   }}void Update(int l ,int r ,int t ,int a ,int b ,int c){   if(a == l && b == r)   {      cnt[t] += c;      Pushup(l ,r ,t);      return;   }   int mid = (l + r) >> 1;   if(b <= mid) Update(lson ,a ,b ,c);   else if(a >= mid) Update(rson ,a ,b ,c);   else    {      Update(lson ,a ,mid ,c);      Update(rson ,mid ,b ,c);   }   Pushup(l ,r ,t);}int search_2(int id ,double now){   int low = 1 ,up = id ,mid ,ans;   while(low <= up)   {      mid = (low + up) >> 1;      if(now <= num[mid])      {         ans = mid;         up = mid - 1;      }      else low = mid + 1;   }   return ans;}int main (){   int t ,i ,id ,n;   double x1 ,x2 ,y1 ,y2 ,sum;   scanf("%d" ,&t);   while(t--)   {      scanf("%d" ,&n);      for(id = 0 ,i = 1 ;i <= n ;i ++)      {         scanf("%lf %lf %lf %lf" ,&x1 ,&y1 ,&x2 ,&y2);         edge[++id].l = x1;         edge[id].r = x2 ,edge[id].h = y1 ,edge[id].mk = 1;         tmp[id] = x1;                  edge[++id].l = x1;         edge[id].r = x2 ,edge[id].h = y2 ,edge[id].mk = -1;         tmp[id] = x2;      }      sort(tmp + 1 ,tmp + id + 1);      sort(edge + 1 ,edge + id + 1 ,camp);      for(id = 0 ,i = 1 ;i <= n * 2 ;i ++)      if(i == 1 || tmp[i] != tmp[i-1])      num[++id] = tmp[i];      memset(cnt ,0 ,sizeof(cnt));      memset(len1 ,0 ,sizeof(len1));      memset(len2 ,0 ,sizeof(len2));      edge[0].h = edge[1].h;      for(sum = 0 ,i = 1 ;i <= n * 2 ;i ++)      {         sum += len2[1] * (edge[i].h - edge[i-1].h);         int l = search_2(id ,edge[i].l);         int r = search_2(id ,edge[i].r);         Update(1 ,id ,1 ,l ,r ,edge[i].mk);      }      printf("%.2lf\n" ,sum);   }   return 0;}

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