HDU 5324 Boring Class CDQ分治

题目传送门

题目要求一个3维偏序点的最长子序列，并且字典序最小。

题解：

这种题目出现的次数特别多了。如果不需要保证字典序的话直接cdq就好了。

这里需要维护字典序的话，我们从后往前配对就好了，因为越前面的点权重越大。（对于字典序来说）

代码：

 #include<bits/stdc++.h>
 using namespace std;
 #define Fopen freopen("_in.txt","r",stdin); freopen("_out.txt","w",stdout);
 #define LL long long
 #define ULL unsigned LL
 #define fi first
 #define se second
 #define pb push_back
 #define lson l,m,rt<<1
 #define rson m+1,r,rt<<1|1
 #define lch(x) tr[x].son[0]
 #define rch(x) tr[x].son[1]
 #define max3(a,b,c) max(a,max(b,c))
 #define min3(a,b,c) min(a,min(b,c))
 typedef pair<int,int> pll;
 const int inf = 0x3f3f3f3f;
 const LL INF = 0x3f3f3f3f3f3f3f3f;
 const LL mod =  (int)1e9+;
 const int N = 1e5 + ;
 int ans[N];
 int pre[N];
 struct Node{
     int l, r, id;
     bool operator<(const Node x) const{
         if(r != x.r) return r > x.r;
         return id > x.id;
     }
 }A[N], tmp[N];
 int ll[N], lsz;
 int tree[N][];
 void add(int x, int len, int id){
     for(int i = x; i < N; i+=i&(-i)){
         if(len > tree[i][]){
             tree[i][] = len;
             tree[i][] = id;
         }
         else if(len == tree[i][]) tree[i][] = min(tree[i][], id);
     }
 }
 pll query(int x){
     int len = , ret = ;
     for(int i = x; i > ; i -= i &(-i)){
         if(len == tree[i][]) ret = min(ret, tree[i][]);
         else if(len < tree[i][]){
             len = tree[i][];
             ret = tree[i][];
         }
     }
     return pll(len, ret);
 }
 void clear(int x){
     for(int i = x; i < N; i += i&(-i))
         tree[i][] = tree[i][] = ;
 }
 void cdq(int l, int r){
     if(l == r) return ;
     int m = l+r >> ;
     cdq(m+,r);
     for(int i = l; i <= r; i++)
         tmp[i] = A[i];
     sort(tmp+l, tmp+r+);
     for(int i = l; i <= r; i++){
         //cout << tmp[i].id << " " << tmp[i].l << " " << tmp[i].r << endl;
         int id = tmp[i].id;
         if(id > m) add(tmp[i].l, ans[id], id);
         else {
             pll p = query(tmp[i].l);
             int len = p.fi, iid = p.se;
             if(len == ) continue;
             if(len+ > ans[id]){
                 ans[id] = len+;
                 pre[id] = iid;
             }
             else if(len +  == ans[id])
                 pre[id] = min(pre[id], iid);
         }
     }
     for(int i = l; i <= r; i++){
         if(tmp[i].id > m) clear(tmp[i].l);
     }
     cdq(l,m);
 }
 int main(){
     int n;
         lsz = ;
     while(~scanf("%d", &n)){
         for(int i = ; i <= n; i++){
             ans[i] = , pre[i] = i;
             A[i].id = i;
             scanf("%d", &A[i].l);
             ll[i] = A[i].l;
         }
         sort(ll+, ll++n);
         lsz = unique(ll+,ll++n) - ll - ;
         for(int i = ; i <= n; i++){
             scanf("%d", &A[i].r);
             A[i].l = lower_bound(ll+, ll++lsz, A[i].l) - ll;
         }
         cdq(,n);
         int fans = , fid = ;
         for(int i = ; i <= n; i++){
             if(fans < ans[i]){
                 fans = ans[i];
                 fid = i;
             }
         }
         printf("%d\n", fans);
         for(int i = ; i <= fans; i++){
             printf("%d%c", fid, " \n"[i==fans]);
             fid = pre[fid];
         }

     }
     return ;
 }