hdu 3265 线段树扫描线（拆分矩形）

题意：
       给你n个矩形，每个矩形上都有一个矩形的空洞，所有的矩形都是平行于x,y轴的，最后问所有矩形的覆盖面积是多少。


思路：

      是典型的矩形覆盖问题，只不过每个矩形上多了一个矩形洞，我的做法是吧当前的矩形分成四个小的矩形，然后用线段树的扫描线扫一遍就ok了，记得要用INT64 ,自己没注意这个问题wa了一次。

#include<stdio.h>
#include<string.h>
#include<algorithm>

#define N 300000
#define lson l ,mid ,t << 1
#define rson mid ,r ,t << 1 | 1

using namespace std;

typedef struct
{
   __int64 l ,r ,h ,mk;
}EDGE;

__int64 len[N] ,cnt[N];
EDGE edge[N*2];

bool camp(EDGE a ,EDGE b)
{
   return a.h < b.h || a.h == b.h && a.mk > b.mk;
}

void Pushup(__int64 l ,__int64 r ,__int64 t)
{
   if(cnt[t]) len[t] = r - l;
   else if(l + 1 == r) len[t] = 0;
   else len[t] = len[t<<1] + len[t<<1|1];
}

void Update(__int64 l ,__int64 r ,__int64 t ,__int64 a ,__int64 b ,__int64 c)
{
   //printf("%d %d %d
" ,l ,r ,t);
   if(l == a && r == b)
   {
      cnt[t] += c;
      Pushup(l ,r ,t);
      return;
   }
   __int64 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);
}

__int64 abss(__int64 x)
{
   return x < 0 ? -x : x;
}

int main ()
{
   __int64 i ,j ,n ,x1 ,x2 ,x3 ,x4 ,y1 ,y2 ,y3 ,y4 ,id;
   __int64 xx1 ,xx2 ,yy1 ,yy2;
   while(~scanf("%I64d" ,&n) && n)
   {
      for(id = 0 ,i = 1 ;i <= n ;i ++)
      {
         scanf("%I64d %I64d %I64d %I64d %I64d %I64d %I64d %I64d" ,&x1 ,&y1 ,&x2 ,&y2 ,&x3 ,&y3 ,&x4 ,&y4);
         x1 ++ ,y1 ++ ,x2 ++ ,y2 ++ ,x3 ++ ,y3 ++ ,x4 ++ ,y4 ++;
         // x1 y2 x2 y4
         xx1 = x1 ,xx2 = x2 ,yy1 = y2 ,yy2 = y4;
         if(abss(xx1 - xx2) && abss(yy1 - yy2))
         {
            edge[++id].l = xx1;
            edge[id].r = xx2 ,edge[id].h = yy1 ,edge[id].mk = 1;
            
            edge[++id].l = xx1;
            edge[id].r = xx2 ,edge[id].h = yy2 ,edge[id].mk = -1;
         }
          // x1 y3 x2 y1
         xx1 = x1 ,xx2 = x2 ,yy1 = y3 ,yy2 = y1;
         if(abss(xx1 - xx2) && abss(yy1 - yy2))
         {
            edge[++id].l = xx1;
            edge[id].r = xx2 ,edge[id].h = yy1 ,edge[id].mk = 1;
            
            edge[++id].l = xx1;
            edge[id].r = xx2 ,edge[id].h = yy2 ,edge[id].mk = -1;
         }
         
          // x1 y4 x3 y3
         xx1 = x1 ,xx2 = x3 ,yy1 = y4 ,yy2 = y3;
         if(abss(xx1 - xx2) && abss(yy1 - yy2))
         {
            edge[++id].l = xx1;
            edge[id].r = xx2 ,edge[id].h = yy1 ,edge[id].mk = 1;
            
            edge[++id].l = xx1;
            edge[id].r = xx2 ,edge[id].h = yy2 ,edge[id].mk = -1;
         }
         
          // x4 y4 x2 y3
         xx1 = x4 ,xx2 = x2 ,yy1 = y4 ,yy2 = y3;
         if(abss(xx1 - xx2) && abss(yy1 - yy2))
         {
            edge[++id].l = xx1;
            edge[id].r = xx2 ,edge[id].h = yy1 ,edge[id].mk = 1;
            
            edge[++id].l = xx1;
            edge[id].r = xx2 ,edge[id].h = yy2 ,edge[id].mk = -1;
         }  
      }
      sort(edge + 1 ,edge + id + 1 ,camp);
      __int64 Ans = 0;
      memset(len ,0 ,sizeof(len));
      memset(cnt ,0 ,sizeof(cnt));
      edge[0].h = edge[1].h;
      for(i = 1 ;i <= id ;i ++)
      {
         Ans += len[1] * (edge[i].h - edge[i-1].h);
         Update(1 ,50001,1 ,edge[i].l ,edge[i].r ,edge[i].mk);
         
      }
      printf("%I64d
" ,Ans);
   }
   return 0;
}