K-th Number POJ - 2104 （整体二分）

Posted 2020-10-21 yijiull

K-th Number

 POJ - 2104 

 

之前学主席树写了一遍

最近再看CDQ分治和整体二分，一直不是很理解，看着别人代码稍微理解了一些

 1 //比主席树慢了挺多
 2 #include <iostream>
 3 #include <cstring>
 4 #include <cstdio>
 5 
 6 using namespace std;
 7 
 8 const int maxn = 1e5 + 10;
 9 const int maxq = 5010;
10 const int inf = 0x3f3f3f3f;
11 
12 struct Qry{
13     int x, y, k;
14     int id, type;
15     Qry(int x = 0, int y = 0, int k = 0, int id = 0, int type = 0):
16         x(x), y(y), k(k), id(id), type(type){}
17 }q[maxn + maxq], q1[maxn + maxq], q2[maxn + maxq];
18 
19 struct Bit{
20     int n;
21     int a[maxn];
22     void init(int _n){
23         n = _n;
24         memset(a, 0, sizeof a);
25     }
26     void add(int i, int x){
27         for(; i <= n; i += i & -i) a[i] += x;
28     }
29     int sum(int i){
30         int res = 0;
31         for(; i; i -= i & -i) res += a[i];
32         return res;
33     }
34 }bit;
35 
36 int n, m, a[maxn];
37 int ans[maxq];
38 
39 void solve(int L, int R, int l, int r){
40     if(L > R) return;
41     if(l == r){
42         for(int i = L; i <= R; i++){
43             if(q[i].type == 2) ans[q[i].id] = l;
44         }
45         return;
46     }
47     int m = l + r >> 1;
48     int f = 0, g = 0;
49     for(int i = L; i <= R; i++){
50         if(q[i].type == 1){
51             if(q[i].x <= m) {
52                 bit.add(q[i].id, 1);
53                 q1[f++] = q[i];
54             }else q2[g++] = q[i];
55         }else{
56             int temp = bit.sum(q[i].y) - bit.sum(q[i].x - 1);
57             if(temp >= q[i].k) {
58                 q1[f++] = q[i];
59             }else{
60                 q[i].k -= temp;
61                 q2[g++] = q[i];
62             }
63         }
64     }
65     for(int i = 0; i < f; i++) if(q1[i].type == 1) bit.add(q1[i].id, -1);
66     memcpy(q + L, q1, f * sizeof(Qry));
67     memcpy(q + L + f, q2, g * sizeof(Qry));
68     solve(L, L + f - 1, l, m);
69     solve(L + f, R, m + 1, r);
70 }
71 int main(){
72     ios::sync_with_stdio(0);
73     while(cin>>n>>m){
74         bit.init(n);
75         int idx = 0;
76         for(int i = 1; i <= n; i++){
77             cin>>a[i];
78             q[++idx] = Qry(a[i], 0, 0, i, 1);
79         }
80         for(int i = 1; i <= m; i++){
81             int x, y, k;
82             cin>>x>>y>>k;
83             q[++idx] = Qry(x, y, k, i, 2);
84         }
85         solve(1, idx, -inf, inf);
86         for(int i = 1; i <= m; i++){
87             cout<<ans[i]<<endl;
88         }
89     }
90 }

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