标签：namespace 规模 tin 输出 cond [1] class amp 数据
[codeforces722D]Generating Sets
试题描述
You are given a set Y of n distinct positive integers y1,?y2,?...,?yn.
Set X of n distinct positive integers x1,?x2,?...,?xn is said to generate set Y if one can transform X to Y by applying some number of the following two operation to integers in X:
Note that integers in X are not required to be distinct after each operation.
Two sets of distinct integers X and Y are equal if they are equal as sets. In other words, if we write elements of the sets in the array in the increasing order, these arrays would be equal.
Note, that any set of integers (or its permutation) generates itself.
You are given a set Y and have to find a set X that generates Y and the maximum element of X is mininum possible.
输入
The first line of the input contains a single integer n (1?≤?n?≤?50?000) — the number of elements in Y.
The second line contains n integers y1,?...,?yn (1?≤?yi?≤?109), that are guaranteed to be distinct.
输出
Print n integers — set of distinct integers that generate Y and the maximum element of which is minimum possible. If there are several such sets, print any of them.
输入示例
6 9 7 13 17 5 11
输出示例
4 5 2 6 3 1
数据规模及约定
见“输入”
题解
建一颗二叉的 Trie 树，把每个数转化成二进制并映射到树上，然后每次找到最大的那个数往根节点上跳：一步步向上走，一遇到没有被占用的点就停下来，定居在这里；如果一直到根节点都是被别人占用的点，就说明不可能再把最大值减小了，输出并退出程序即可。
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <stack>
#include <vector>
#include <queue>
#include <cstring>
#include <string>
#include <map>
#include <set>
using namespace std;
const int BufferSize = 1 << 16;
char buffer[BufferSize], *Head, *Tail;
inline char Getchar() {
if(Head == Tail) {
int l = fread(buffer, 1, BufferSize, stdin);
Tail = (Head = buffer) + l;
}
return *Head++;
}
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 50010
#define maxnode 1600010
#define maxlog 32
int rt, ToT, fa[maxnode], ch[maxnode][2], node[maxn], id2[maxlog];
bool tag[maxnode];
void build(int& o, int v) {
int num[maxlog], cnt = 0;
while(v) num[++cnt] = v & 1, v >>= 1;
// for(int i = cnt; i; i--) putchar(num[i] + ‘0‘); putchar(‘\n‘);
if(!o) o = rt;
for(int i = cnt - 1; i; i--) {
int& to = ch[o][num[i]];
if(!to) to = ++ToT, fa[to] = o;
o = to;
}
tag[o] = 1;
return ;
}
int Move(int& o) {
int u = o;
while(fa[u] && tag[u]) u = fa[u];
if(u && !tag[u]) tag[o] = 0, tag[u] = 1, o = u;
int res = 1; u = o;
int num[maxlog], cnt = 0;
while(fa[u]) num[++cnt] = (ch[fa[u]][1] == u), u = fa[u];
for(int i = cnt; i; i--) res = res << 1 | num[i];
return res;
}
struct Num {
int v, id;
Num() {}
Num(int _, int __): v(_), id(__) {}
bool operator < (const Num& t) const { return v < t.v; }
} ;
priority_queue <Num> Q;
int ans[maxn], cnta;
int main() {
id2[0] = 1;
for(int i = 1; i < maxlog; i++) id2[i] = id2[i-1] << 1;
rt = ToT = 1;
int n = read();
for(int i = 1; i <= n; i++) {
int x = read(); Q.push(Num(x, i));
build(node[i], x);
}
/*for(int i = 1; i <= ToT; i++) printf("%d: %d(%d %d)[%d]\n", i, fa[i], ch[i][0], ch[i][1], tag[i]);
for(int i = 1; i <= n; i++) printf("%d%c", node[i], i < n ? ‘ ‘: ‘\n‘);*/
while(1) {
Num u = Q.top(); Q.pop();
int tmp = Move(node[u.id]);
// printf("%d -> %d(%d)\n", u.v, tmp, node[u.id]);
Q.push(Num(tmp, u.id));
if(tmp == u.v) break;
}
while(!Q.empty()) ans[++cnta] = Q.top().v, Q.pop();
for(int i = 1; i <= cnta; i++) printf("%d%c", ans[i], i < cnta ? ‘ ‘ : ‘\n‘);
return 0;
}
[codeforces722D]Generating Sets
标签：namespace 规模 tin 输出 cond [1] class amp 数据
原文地址：http://www.cnblogs.com/xiao-ju-ruo-xjr/p/6349343.html
