[BZOJ1901]Zju2112 Dynamic Rankings
试题描述
输入
输出
输入示例
5 3 3 2 1 4 7 Q 1 4 3 C 2 6 Q 2 5 3
输出示例
3 6
数据规模及约定
见“试题描述”
题解
打了一波线段树套 treap,相比于主席树套树状数组来说慢多了。。。
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cctype>
#include <algorithm>
using namespace std;
int read() {
int x = 0, f = 1; char c = getchar();
while(!isdigit(c)){ if(c == ‘-‘) f = -1; c = getchar(); }
while(isdigit(c)){ x = x * 10 + c - ‘0‘; c = getchar(); }
return x * f;
}
#define maxn 10005
#define maxnode 1280010
struct Node {
int v, r, siz;
Node() {}
Node(int _, int __): v(_), r(__) {}
} ns[maxnode];
int ToT, rt[maxn<<2], fa[maxnode], ch[2][maxnode];
void maintain(int o) {
ns[o].siz = 1;
for(int i = 0; i < 2; i++) if(ch[i][o])
ns[o].siz += ns[ch[i][o]].siz;
return ;
}
void rotate(int u) {
int y = fa[u], z = fa[y], l = 0, r = 1;
if(z) ch[ch[1][z]==y][z] = u;
if(ch[1][y] == u) swap(l, r);
fa[u] = z; fa[y] = u; fa[ch[r][u]] = y;
ch[l][y] = ch[r][u]; ch[r][u] = y;
maintain(y); maintain(u);
return ;
}
void insert(int& o, int v) {
if(!o) {
ns[o = ++ToT] = Node(v, rand());
return maintain(o);
}
bool d = v > ns[o].v;
insert(ch[d][o], v); fa[ch[d][o]] = o;
if(ns[ch[d][o]].r > ns[o].r) {
int t = ch[d][o];
rotate(t); o = t;
}
return maintain(o);
}
void del(int& o, int v) {
if(!o) return ;
if(ns[o].v == v) {
if(!ch[0][o] && !ch[1][o]) o = 0;
else if(!ch[0][o]) {
int t = ch[1][o]; fa[t] = fa[o]; o = t;
}
else if(!ch[1][o]) {
int t = ch[0][o]; fa[t] = fa[o]; o = t;
}
else {
bool d = ns[ch[1][o]].r > ns[ch[0][o]].r;
int t = ch[d][o]; rotate(t); o = t;
del(ch[d^1][o], v);
}
}
else {
bool d = v > ns[o].v;
del(ch[d][o], v);
}
return maintain(o);
}
int Find(int o, int v) {
if(!o) return 0;
int ls = ch[0][o] ? ns[ch[0][o]].siz : 0;
if(v >= ns[o].v) return ls + 1 + Find(ch[1][o], v);
return Find(ch[0][o], v);
}
int n, val[maxn];
void build(int L, int R, int o, int p) {
insert(rt[o], val[p]);
if(L == R) return ;
int M = L + R >> 1, lc = o << 1, rc = lc | 1;
if(p <= M) build(L, M, lc, p);
else build(M+1, R, rc, p);
return ;
}
void update(int L, int R, int o, int p, int v) {
del(rt[o], val[p]); insert(rt[o], v);
if(L == R) return ;
int M = L + R >> 1, lc = o << 1, rc = lc | 1;
if(p <= M) update(L, M, lc, p, v);
else update(M+1, R, rc, p, v);
return ;
}
int ql, qr;
int query(int L, int R, int o, int v) {
if(ql <= L && R <= qr) return Find(rt[o], v);
int M = L + R >> 1, lc = o << 1, rc = lc | 1, ans = 0;
if(ql <= M) ans += query(L, M, lc, v);
if(qr > M) ans += query(M+1, R, rc, v);
return ans;
}
int main() {
n = read(); int q = read();
for(int i = 1; i <= n; i++) val[i] = read(), build(1, n, 1, i);
while(q--) {
char tp[2]; scanf("%s", tp);
if(tp[0] == ‘Q‘) {
ql = read(); qr = read(); int k = read();
int l = 0, r = (int)1e9; bool has = 0;
while(l < r) {
int mid = l + r >> 1;
if(query(1, n, 1, mid) < k) {
l = mid + 1;
if(query(1, n, 1, l) >= k){ has = 1; printf("%d\n", l); break; }
}
else r = mid;
}
if(!has) printf("%d\n", l);
}
else {
int p = read(), v = read();
update(1, n, 1, p, v);
val[p] = v;
}
}
return 0;
}
[BZOJ1901]Zju2112 Dynamic Rankings
原文:http://www.cnblogs.com/xiao-ju-ruo-xjr/p/6155889.html