BOI2007 Mokia

传送门

这个题其实和板子题差不多……不过这次修改和询问是分离开的，然后一个询问要被拆分成4个，就是维护二维前缀和的方式。

我的做法比较奇怪……（怎么我\(CDQ\)分治的题做法都很奇怪），别人都是按时间排序然后归并x，树状数组统计y，用x作为限制，因为这样时间天然有序。我是按x排序然后归并时间……其实这样也能做，但是一来比较慢（一开始要\(sort\)），二来我非常智障的把每个询问的时间没有搞成同一个值……

之后套板子做就可以了。

// luogu-judger-enable-o2
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<iostream>
#include<cmath>
#include<set>
#include<vector>
#include<map>
#include<queue>
#define rep(i,a,n) for(int i = a;i <= n;i++)
#define per(i,n,a) for(int i = n;i >= a;i--)
#define enter putchar('\n')
#define fr friend inline
#define y1 poj
#define mp make_pair
#define pr pair<int,int>
#define fi first
#define sc second
#define pb push_back
#define lowbit(x) x & (-x)
#define B printf("Bug\n");

using namespace std;
typedef long long ll;
const int M = 400005;
const int N = 2000005;
const double INF = 1e15;
const double eps = 1e-7;

int read()
{
    int ans = 0,op = 1;
    char ch = getchar();
    while(ch < '0' || ch > '9')
    {
    if(ch == '-') op = -1;
    ch = getchar();
    }
    while(ch >= '0' && ch <= '9')
    {
    ans *= 10;
    ans += ch - '0';
    ch = getchar();
    }
    return ans * op;
}

struct node
{
   int x,y,tim,val;
   bool opa;
   bool operator < (const node &g)const
   {
      if(x == g.x && tim == g.tim) return y < g.y;
      if(x == g.x) return tim < g.tim;
      return x < g.x;
   }
}a[M];

int n,cnt,x1,y1,x2,y2,A,op,ans[N],Ti;
bool vis[N];

struct treearray
{
   int c[N];
   void add(int x,int v) {while(x <= N-2) c[x] += v,x += lowbit(x);}
   int ask(int x) {int cur = 0;while(x) cur += c[x],x -= lowbit(x); return cur;}
}T;

bool cmp(const node &p,const node &q)
{
   if(p.tim == q.tim) return p.y < q.y;
   return p.tim < q.tim;
}
   
void CDQ(const int &l,const int &r)
{
   if(l == r) return;
   int mid = (l+r) >> 1;
   CDQ(l,mid),CDQ(mid+1,r);
   sort(a+l,a+mid+1,cmp),sort(a+mid+1,a+r+1,cmp);
   int j = l;
   rep(i,mid+1,r)
   {
      if(a[i].opa) continue;
      while(j <= mid && a[j].tim < a[i].tim)
      {
     if(!a[j].opa) {j++;continue;}
     T.add(a[j].y,a[j].val),j++;
      }
      ans[a[i].tim] += a[i].val * T.ask(a[i].y);
   }
   rep(i,l,j-1) if(a[i].opa) T.add(a[i].y,-a[i].val);
}

int main()
{
   n = read(),n = read();
   while(scanf("%d",&op) && op != 3)
   {
      Ti++;
      if(op == 1) a[++cnt].x = read(),a[cnt].y = read(),a[cnt].val = read(),a[cnt].opa = 1,a[cnt].tim = Ti;
      else
      {
     x1 = read(),y1 = read(),x2 = read(),y2 = read(),vis[Ti] = 1;
     a[++cnt].x = x1-1,a[cnt].y = y1-1,a[cnt].val = 1,a[cnt].tim = Ti;
     a[++cnt].x = x1-1,a[cnt].y = y2,a[cnt].val = -1,a[cnt].tim = Ti;
     a[++cnt].x = x2,a[cnt].y = y1-1,a[cnt].val = -1,a[cnt].tim = Ti;
     a[++cnt].x = x2,a[cnt].y = y2,a[cnt].val = 1,a[cnt].tim = Ti;
      }
   }
   sort(a+1,a+1+cnt);
   CDQ(1,cnt);
   rep(i,1,Ti) if(vis[i]) printf("%d\n",ans[i]);
   return 0;
}