[POJ2104]K-th Number（主席树，静态区间k小）

题目链接：http://poj.org/problem?id=2104

题意：给一个数列，求给定区间第k小的数是多少。

思路：可以按照数字出现的次数为值建立线段树，每插入一个数字就维护一棵线段树，这样求某个区间(比如求[x,y]区间内)的第k小时就可以二分了。当tree[tree[x].l].sum - tree[tree[y].l].sum >= k时去查左儿子，tree[tree[x].l].sum - tree[tree[y].l].sum < k时去查右儿子。数据大，所以要离散化后再做。

  1 /*
  2 ━━━━━┒ギリギリ♂ eye！
  3 ┓┏┓┏┓┃キリキリ♂ mind！
  4 ┛┗┛┗┛┃＼○／
  5 ┓┏┓┏┓┃ /
  6 ┛┗┛┗┛┃ノ)
  7 ┓┏┓┏┓┃
  8 ┛┗┛┗┛┃
  9 ┓┏┓┏┓┃
 10 ┛┗┛┗┛┃
 11 ┓┏┓┏┓┃
 12 ┛┗┛┗┛┃
 13 ┓┏┓┏┓┃
 14 ┃┃┃┃┃┃
 15 ┻┻┻┻┻┻
 16 */
 17 #include <algorithm>
 18 #include <iostream>
 19 #include <iomanip>
 20 #include <cstring>
 21 #include <climits>
 22 #include <complex>
 23 #include <fstream>
 24 #include <cassert>
 25 #include <cstdio>
 26 #include <bitset>
 27 #include <vector>
 28 #include <deque>
 29 #include <queue>
 30 #include <stack>
 31 #include <ctime>
 32 #include <set>
 33 #include <map>
 34 #include <cmath>
 35 using namespace std;
 36 #define fr first
 37 #define sc second
 38 #define cl clear
 39 #define BUG puts("here!!!")
 40 #define W(a) while(a--)
 41 #define pb(a) push_back(a)
 42 #define Rint(a) scanf("%d", &a)
 43 #define Rll(a) scanf("%I64d", &a)
 44 #define Rs(a) scanf("%s", a)
 45 #define Cin(a) cin >> a
 46 #define FRead() freopen("in", "r", stdin)
 47 #define FWrite() freopen("out", "w", stdout)
 48 #define Rep(i, len) for(int i = 0; i < (len); i++)
 49 #define For(i, a, len) for(int i = (a); i < (len); i++)
 50 #define Cls(a) memset((a), 0, sizeof(a))
 51 #define Clr(a, x) memset((a), (x), sizeof(a))
 52 #define Full(a) memset((a), 0x7f7f7f, sizeof(a))
 53 #define lrt rt << 1
 54 #define rrt rt << 1 | 1
 55 #define pi 3.14159265359
 56 #define RT return
 57 #define lowbit(x) x & (-x)
 58 #define onecnt(x) __builtin_popcount(x)
 59 typedef long long LL;
 60 typedef long double LD;
 61 typedef unsigned long long ULL;
 62 typedef pair<int, int> pii;
 63 typedef pair<string, int> psi;
 64 typedef pair<LL, LL> pll;
 65 typedef map<string, int> msi;
 66 typedef vector<int> vi;
 67 typedef vector<LL> vl;
 68 typedef vector<vl> vvl;
 69 typedef vector<bool> vb;
 70
 71 const int maxn = 1000010;
 72 int n, m;
 73 int cnt, root[maxn], a[maxn], x, y, k;
 74 typedef struct Node {
 75     int l, r, sum;
 76 }Node;
 77 int h[maxn], hcnt;
 78 Node tree[maxn*40];
 79
 80 int getid(int x) {
 81     return lower_bound(h+1, h+hcnt+1, x) - h;
 82 }
 83
 84 int build(int l, int r) {
 85     int rt = cnt++;
 86     tree[rt].sum = 0;
 87     if(l != r) {
 88         int mid = (l + r) >> 1;
 89         tree[rt].l = build(l, mid);
 90         tree[rt].r = build(mid+1, r);
 91     }
 92     return rt;
 93 }
 94
 95 int update(int rt, int pos, int val) {
 96     int k = cnt++, tmp = k;
 97     tree[k].sum = tree[rt].sum + val;
 98     int l = 1, r = hcnt;
 99     while(l < r) {
100         int mid = (l + r) >> 1;
101         if(pos <= mid) {
102             tree[k].l = cnt++; tree[k].r = tree[rt].r;
103             k = tree[k].l; rt = tree[rt].l;
104             r = mid;
105         }
106         else {
107             tree[k].r = cnt++; tree[k].l = tree[rt].l;
108             k = tree[k].r; rt = tree[rt].r;
109             l = mid + 1;
110         }
111         tree[k].sum = tree[rt].sum + val;
112     }
113     return tmp;
114 }
115
116 int query(int x, int y, int k) {
117     int l = 1, r = hcnt;
118     while(l < r) {
119         int mid = (l + r) >> 1;
120         if(tree[tree[x].l].sum - tree[tree[y].l].sum >= k) {
121             r = mid;
122             x = tree[x].l;
123             y = tree[y].l;
124         }
125         else {
126             l = mid + 1;
127             k -= tree[tree[x].l].sum - tree[tree[y].l].sum;
128             x = tree[x].r;
129             y = tree[y].r;
130         }
131     }
132     return l;
133 }
134
135 int main() {
136     // FRead();
137     cnt = 0;
138     Rint(n); Rint(m);
139     For(i, 1, n+1) {
140         Rint(a[i]);
141         h[i] = a[i];
142     }
143     sort(h+1, h+n+1);
144     hcnt = unique(h+1, h+n+1) - h - 1;
145     root[n+1] = build(1, hcnt);
146     for(int i = n; i >= 1; i--) {
147         root[i] = update(root[i+1], getid(a[i]), 1);
148     }
149     W(m) {
150         Rint(x), Rint(y), Rint(k);
151         printf("%d\n", h[query(root[x], root[y+1], k)]);
152     }
153     return 0;
154 }