bzoj 4237稻草人

　　按x轴进行分治，将[l,r]分成[l,mid]和[mid+1,r]，左下角点x值在[l,mid]中，右上角点x值在[mid+1,r],然后将[l,r]中的所有点按y轴排序，按顺序扫描，若扫描到左下角点，用一个单调栈维护，若扫描到右上角点，用另一个单调栈维护的同时，去维护左下角的单调栈中二分出答案，复杂度O(nlognlogn),程序跑的有点慢

　　代码

 #include<cstdio>
 #include<algorithm>
 #include<set>
 #define N 500010
 using namespace std;
 int n,i;
 long long ans;
 int top1,top2,stack1[N],stack2[N];
 struct g{
     int x,y;
 }a[N];
 bool cmp(g a,g b)
 {
     return a.x<b.x;
 }
 bool cmp1(g a,g b)
 {
     return a.y<b.y;
 }
 int ef(int x)
 {
     int l=,r=top1,m;
     while (l<=r)
     {
         m=(l+r)>>;
         if (x>a[stack1[m]].y) l=m+;else r=m-;
     }
     return r;
 }
 void solve(int L,int R,int l,int r)
 {
     int p,q;
     if (L>=R) return;
     if (l>=r) return;
     int m=(L+R)>>;
     int cnt=l,i;
     int t;
     for (i=l;i<=r;i++)
     if (a[i].x<=m)
     {
         t=a[i].x;a[i].x=a[cnt].x;a[cnt].x=t;
         t=a[i].y;a[i].y=a[cnt].y;a[cnt].y=t;
         cnt++;
     }    

     if ((l<cnt)&&(cnt<=r))
     {
         sort(a+l,a+cnt,cmp1);
         sort(a+cnt,a++r,cmp1);

         top1=;top2=;
         p=l;q=cnt;
         while ((p<cnt)||(q<=r))
         {
             if ((p==cnt)||((q<=r)&&(a[q].y<a[p].y)))
             {
                 while ((top2)&&(a[q].x<a[stack2[top2]].x)) top2--;
                 ans+=ef(a[q].y)-ef(a[stack2[top2]].y);
                 top2++;stack2[top2]=q;q++;
             }
             else
             {
                 while ((top1)&&(a[p].x>a[stack1[top1]].x)) top1--;
                 top1++;stack1[top1]=p;p++;
             }
         }

     }
     /*
     printf("%d %d\n",L,R);
     printf("A:\n");
     for (i=l;i<cnt;i++)
     printf("%d %d\n",a[i].x,a[i].y);
         printf("B:\n");
     for (i=cnt;i<=r;i++)
     printf("%d %d\n",a[i].x,a[i].y);
      */
     solve(L,m,l,cnt-);
     solve(m+,R,cnt,r);
 }
 int main()
 {
     scanf("%d",&n);
     for (i=;i<=n;i++)
     {
         scanf("%d%d",&a[i].x,&a[i].y);
         a[i].y++;
     }
     sort(a+,a++n,cmp);
     for (i=;i<=n;i++)
     a[i].x=i;
     solve(,n,,n);
     printf("%lld\n",ans);
 }