SPOJ:Another Longest Increasing Subsequence Problem(CDQ分治求三维偏序)

Given a sequence of N pairs of integers, find the length of the longest increasing subsequence of it.

An increasing sequence A1..An is a sequence such that for every i < j, Ai < Aj.

A subsequence of a sequence is a sequence that appears in the same relative order, but not necessarily contiguous.

A pair of integers (x1, y1) is less than (x2, y2) iff x1 < x2 and y1 < y2.

Input

The first line of input contains an integer N (2 ≤ N ≤ 100000).

The following N lines consist of N pairs of integers (xi, yi) (-109 ≤ xi, yi ≤ 109).

Output

The output contains an integer: the length of the longest increasing subsequence of the given sequence.

Example

Input:
8
1 3
3 2
1 1
4 5
6 3
9 9
8 7
7 6

Output:
3

题意；求最长的序列LIS，满足i<j，xi<xj ；yi<yj。

思路：裸题，cqd分治，计算每个[L,Mid]区间对[Mid+1,R]区间的贡献。

有两个注意点：

第一：由于时间紧，只有300ms，不能写结构体的排序； 这里借鉴了别人的一维排序（ORZ，强的啊）。

第二：注意规避x1=x2，y1<y2的时候不能用 1去更新2。（和求逆序对那题一样，只有把y坐标的放左边即可）。

#include<bits/stdc++.h>
using namespace std;
const int maxn=1000010;
int p[maxn],a[maxn],b[maxn],dp[maxn],Mx[maxn],tot,ans;
bool cmp(int x,int y){ if(a[x]==a[y]) return x>y; return a[x]<a[y]; }
void solve(int L,int R)
{
if(L==R){ dp[L]=max(dp[L],1); return ;}
int Mid=(L+R)/2;
solve(L,Mid);
for(int i=L;i<=R;i++) p[i]=i;
sort(p+L,p+R+1,cmp);
for(int i=L;i<=R;i++){
if(p[i]<=Mid) for(int j=b[p[i]];j<=tot;j+=(-j)&j) Mx[j]=max(Mx[j],dp[p[i]]);
else for(int j=b[p[i]]-1;j;j-=(-j)&j) dp[p[i]]=max(dp[p[i]],Mx[j]+1);
}
for(int i=L;i<=R;i++)
if(p[i]<=Mid) for(int j=b[p[i]];j<=tot;j+=(-j)&j) Mx[j]=0;
solve(Mid+1,R);
}
int main()
{
int N,i,fcy=0;
scanf("%d",&N);
for(i=1;i<=N;i++) scanf("%d%d",&a[i],&b[i]),p[i]=b[i];
sort(p+1,p+N+1);
tot=unique(p+1,p+N+1)-(p+1);
for(i=1;i<=N;i++) b[i]=lower_bound(p+1,p+tot+1,b[i])-p;
solve(1,N);
for(i=1;i<=N;i++) fcy=max(fcy,dp[i]);
printf("%d\n",fcy);
return 0;
}