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  • 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:

      1 #include<bits/stdc++.h>
      2 using namespace std;
      3 #define Fopen freopen("_in.txt","r",stdin); freopen("_out.txt","w",stdout);
      4 #define LL long long
      5 #define ULL unsigned LL
      6 #define fi first
      7 #define se second
      8 #define pb push_back
      9 #define lson l,m,rt<<1
     10 #define rson m+1,r,rt<<1|1
     11 #define lch(x) tr[x].son[0]
     12 #define rch(x) tr[x].son[1]
     13 #define max3(a,b,c) max(a,max(b,c))
     14 #define min3(a,b,c) min(a,min(b,c))
     15 typedef pair<int,int> pll;
     16 const int inf = 0x3f3f3f3f;
     17 const LL INF = 0x3f3f3f3f3f3f3f3f;
     18 const LL mod =  (int)1e9+7;
     19 const int N = 1e5 + 100;
     20 int ans[N];
     21 int pre[N];
     22 struct Node{
     23     int l, r, id;
     24     bool operator<(const Node x) const{
     25         if(r != x.r) return r > x.r;
     26         return id > x.id;
     27     }
     28 }A[N], tmp[N];
     29 int ll[N], lsz;
     30 int tree[N][2];
     31 void add(int x, int len, int id){
     32     for(int i = x; i < N; i+=i&(-i)){
     33         if(len > tree[i][0]){
     34             tree[i][0] = len;
     35             tree[i][1] = id;
     36         }
     37         else if(len == tree[i][0]) tree[i][1] = min(tree[i][1], id);
     38     }
     39 }
     40 pll query(int x){
     41     int len = 0, ret = 0;
     42     for(int i = x; i > 0; i -= i &(-i)){
     43         if(len == tree[i][0]) ret = min(ret, tree[i][1]);
     44         else if(len < tree[i][0]){
     45             len = tree[i][0];
     46             ret = tree[i][1];
     47         }
     48     }
     49     return pll(len, ret);
     50 }
     51 void clear(int x){
     52     for(int i = x; i < N; i += i&(-i))
     53         tree[i][0] = tree[i][1] = 0;
     54 }
     55 void cdq(int l, int r){
     56     if(l == r) return ;
     57     int m = l+r >> 1;
     58     cdq(m+1,r);
     59     for(int i = l; i <= r; i++)
     60         tmp[i] = A[i];
     61     sort(tmp+l, tmp+r+1);
     62     for(int i = l; i <= r; i++){
     63         //cout << tmp[i].id << " " << tmp[i].l << " " << tmp[i].r << endl;
     64         int id = tmp[i].id;
     65         if(id > m) add(tmp[i].l, ans[id], id);
     66         else {
     67             pll p = query(tmp[i].l);
     68             int len = p.fi, iid = p.se;
     69             if(len == 0) continue;
     70             if(len+1 > ans[id]){
     71                 ans[id] = len+1;
     72                 pre[id] = iid;
     73             }
     74             else if(len + 1 == ans[id])
     75                 pre[id] = min(pre[id], iid);
     76         }
     77     }
     78     for(int i = l; i <= r; i++){
     79         if(tmp[i].id > m) clear(tmp[i].l);
     80     }
     81     cdq(l,m);
     82 }
     83 int main(){
     84     int n;
     85         lsz = 0;
     86     while(~scanf("%d", &n)){
     87         for(int i = 1; i <= n; i++){
     88             ans[i] = 1, pre[i] = i;
     89             A[i].id = i;
     90             scanf("%d", &A[i].l);
     91             ll[i] = A[i].l;
     92         }
     93         sort(ll+1, ll+1+n);
     94         lsz = unique(ll+1,ll+1+n) - ll - 1;
     95         for(int i = 1; i <= n; i++){
     96             scanf("%d", &A[i].r);
     97             A[i].l = lower_bound(ll+1, ll+1+lsz, A[i].l) - ll;
     98         }
     99         cdq(1,n);
    100         int fans = 0, fid = 0;
    101         for(int i = 1; i <= n; i++){
    102             if(fans < ans[i]){
    103                 fans = ans[i];
    104                 fid = i;
    105             }
    106         }
    107         printf("%d
    ", fans);
    108         for(int i = 1; i <= fans; i++){
    109             printf("%d%c", fid, " 
    "[i==fans]);
    110             fid = pre[fid];
    111         }
    112 
    113     }
    114     return 0;
    115 }
    View Code
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  • 原文地址:https://www.cnblogs.com/MingSD/p/9857114.html
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