bzoj3262: 陌上花开

链接：http://www.lydsy.com/JudgeOnline/problem.php?id=3262

题意：中文题。

分析：cdq分治练习题，详细分析戳这里，不过这题没有好的时间戳z，因为这题的三个值都是在1~k范围内的，那就是说有的值不唯一有的不存在，那么我们在分治的时候就要注意边界了，所以我改成了4个参数的分治，详见代码。O(nlogk^2)。

代码：

#include<map>
#include<set>
#include<cmath>
#include<queue>
#include<bitset>
#include<math.h>
#include<cstdio>
#include<vector>
#include<string>
#include<cstring>
#include<iostream>
#include<algorithm>
#pragma comment(linker, "/STACK:102400000,102400000")
using namespace std;
const int N=100010;
const int MAX=1000000100;
const int mod=100000000;
const int MOD1=1000000007;
const int MOD2=1000000009;
const double EPS=0.00000001;
typedef long long ll;
const ll MOD=998244353;
const ll INF=10000000010;
typedef double db;
typedef unsigned long long ull;
struct node {
int x,y,z,id;
}a
,b
;
int k,f[2*N],ans
,p
;
int cmp(node a,node b) {
if (a.x!=b.x) return a.x<b.x;
else if (a.y!=b.y) return a.y<b.y;
else if (a.z!=b.z) return a.z<b.z;
else return a.id<b.id;
}
void add(int a,int b) {
for (;a<=k;a+=a&(-a)) f[a]+=b;
}
int getsum(int a) {
int ret=0;
for (;a;a-=a&(-a)) ret+=f[a];
return ret;
}
void cdq(int l,int r,int L,int R) {
if (l>=r) return ;
int i,mid=(L+R)>>1,w=l,z;
if (L==R) {
for (i=l;i<=r;i++) { ans[a[i].id]+=getsum(a[i].y);add(a[i].y,1); }
for (i=l;i<=r;i++) add(a[i].y,-1);
return ;
}
for (i=l;i<=r;i++)
if (a[i].z<=mid) add(a[i].y,1);
else ans[a[i].id]+=getsum(a[i].y);
for (i=l;i<=r;i++)
if (a[i].z<=mid) add(a[i].y,-1);
for (i=l;i<=r;i++)
if (a[i].z<=mid) { b[w]=a[i];w++; }
z=w-1;
for (i=l;i<=r;i++)
if (a[i].z>mid) { b[w]=a[i];w++; }
for (i=l;i<=r;i++) a[i]=b[i];
cdq(l,z,L,mid);cdq(z+1,r,mid+1,R);
}
int main()
{
int i,n;
scanf("%d%d", &n, &k);
for (i=1;i<=n;i++) {
scanf("%d%d%d", &a[i].x, &a[i].y, &a[i].z);a[i].id=i;
}
sort(a+1,a+n+1,cmp);
a[n+1].x=a[n+1].y=a[n+1].z=0;
memset(f,0,sizeof(f));
memset(ans,0,sizeof(ans));
for (i=n;i;i--)
if (a[i].x==a[i+1].x&&a[i].y==a[i+1].y&&a[i].z==a[i+1].z) {
f[i]=f[i+1]+1;ans[a[i].id]+=f[i];
}
memset(f,0,sizeof(f));
cdq(1,n,1,k);
memset(p,0,sizeof(p));
for (i=1;i<=n;i++) p[ans[i]]++;
for (i=0;i<n;i++) printf("%d\n", p[i]);
return 0;
}