BOI2007 Mokia
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;
}
当你意识到,每个上一秒都成为永恒。