标签:double using ++i build poj inf link print mod
题意:给定一个序列,然后有 q 个询问,每次询问 l - r 区间内的第 k 大的值。
析:很明显的主席树,而且还是裸的主席树,先进行离散化,然后用主席树进行查询就好。
代码如下:
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#include <list>
#define debug() puts("++++");
#define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 100000 + 10;
const int mod = 10;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c) {
return r >= 0 && r < n && c >= 0 && c < m;
}
int a[maxn], t[maxn];
int T[maxn], lch[maxn*30], rch[maxn*30], c[maxn*30];
int tot;
int f(int x){
return lower_bound(t+1, t+m+1, x) - t;
}
int build(int l, int r){
int rt = tot++;
c[rt] = 0;
if(l == r) return rt;
int m = l + r >> 1;
lch[rt] = build(l, m);
rch[rt] = build(m+1, r);
return rt;
}
int update(int rt, int pos, int val){
int newrt = tot++;
int tmp = newrt;
c[newrt] = c[rt] + val;
int l = 1, r = m;
while(l < r){
int m = l + r >> 1;
if(pos <= m){
r = m;
lch[newrt] = tot++;
rch[newrt] = rch[rt];
newrt = lch[newrt];
rt = lch[rt];
}
else{
l = m + 1;
rch[newrt] = tot++;
lch[newrt] = lch[rt];
newrt = rch[newrt];
rt = rch[rt];
}
c[newrt] = c[rt] + val;
}
return tmp;
}
int query(int lrt, int rrt, int k){
int l = 1, r = m;
while(l < r){
int m = l + r >> 1;
if(c[lch[lrt]] - c[lch[rrt]] >= k){
r = m;
lrt = lch[lrt];
rrt = lch[rrt];
}
else{
l = m + 1;
k -= c[lch[lrt]] - c[lch[rrt]];
lrt = rch[lrt];
rrt = rch[rrt];
}
}
return l;
}
int main(){
int q;
while(scanf("%d %d", &n, &q) == 2){
for(int i = 1; i <= n; ++i){
scanf("%d", a+i);
t[i] = a[i];
}
sort(t+1, t+n+1);
m = unique(t+1, t+1+n) - t - 1;
tot = 0;
T[n+1] = build(1, m);
for(int i = n; i; --i){
int pos = f(a[i]);
T[i] = update(T[i+1], pos, 1);
}
while(q--){
int l, r, k;
scanf("%d %d %d", &l, &r, &k);
printf("%d\n", t[query(T[l], T[r+1], k)]);
}
}
return 0;
}
标签:double using ++i build poj inf link print mod
原文地址:http://www.cnblogs.com/dwtfukgv/p/7259039.html