SPOJ - LIS2 Another Longest Increasing Subsequence Problem
cdq分治,dp(i)表示以i为结尾的最长LIS,那么dp的递推是依赖于左边的。
因此在分治的时候需要利用左边的子问题来递推右边。
(345ms? 区间树TLE
/*********************************************************
* ------------------ *
* author AbyssFish *
**********************************************************/
#include<cstdio>
#include<iostream>
#include<string>
#include<cstring>
#include<queue>
#include<vector>
#include<stack>
#include<vector>
#include<map>
#include<set>
#include<algorithm>
#include<cmath>
#include<numeric>
using namespace std; const int MAX_N = +; int dp[MAX_N];
int x[MAX_N], y[MAX_N];
int ys[MAX_N];
int id[MAX_N];
int N; int ns; int *cur;
bool cmp(int a,int b)
{
return cur[a] < cur[b] || (cur[a] == cur[b] && a > b);//这是为了保证严格的单调性
} int compress(int *r, int *dat, int *a, int n)
{
for(int i = ; i < n; i++){
r[i] = i;
}
cur = dat;
sort(r,r+n,cmp);
a[r[]] = ;
for(int i = ; i < n; i++){
int k = r[i], p = r[i-];
a[k] = dat[k] == dat[p]?a[p]:a[p]+;
}
return a[r[n-]];
} int C[MAX_N]; void add(int yi,int d)
{
while(yi <= ns){
C[yi] = max(C[yi],d);
yi += yi&-yi;
}
} int mx_pfx(int yi)
{
int re = ;
while(yi > ){
re = max(C[yi],re);
yi &= yi-;
}
return re;
} void clr(int yi)
{
while(yi <= ns){
C[yi] = ;
yi += yi&-yi;
}
} void dv(int l, int r)
{
if(r-l <= ){
dp[l]++;
}
else {
int md = (l+r)>>;
dv(l,md); for(int i = l; i < r; i++) id[i] = i;
sort(id+l,id+r,cmp); //x维度 for(int i = l; i < r; i++){
int k = id[i];
if(k < md){ //position 维度
add(ys[k],dp[k]); //BIT下标是 y维度
}
else {
//查询位置前保证了BIT里的元素, 位置md之前,x严格小于待查元素
dp[k] = max(dp[k], mx_pfx(ys[k]-));//y严格小于待查元素的最大dp值
}
} for(int i = l; i < r; i++){
if(id[i] < md)
clr(ys[id[i]]);
}
dv(md,r);
}
} void solve()
{
scanf("%d",&N);
for(int i = ; i < N; i++){
scanf("%d%d",x+i,y+i);
}
ns = compress(id,y,ys,N);
cur = x;
dv(,N);
printf("%d\n",*max_element(dp,dp+N));
} //#define LOCAL
int main()
{
#ifdef LOCAL
freopen("in.txt","r",stdin);
#endif
solve();
return ;
}
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