hdu1255 扫描线，矩形重叠面积（两次以上）

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 10000

using 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
" ,sum);
   }
   return 0;
}