Codeforces Round #744 (Div. 3)
比赛链接
Codeforces Round #744 (Div. 3)
E1. Permutation Minimization by Deque
输入 \(n\) 个数，按照输入顺序根据双端队列的插入规则使得字典序最小
解题思路
思维，贪心
贪心策略：按顺序遍历数组元素，模拟双端队列，如果当前元素大于队首元素则放队尾否则放队首
证明：对于当前数，有两种选择，放队尾或队首，按贪心策略得到的部分要比不按贪心策略得到的部分的字典序要小，对于后面的数而言当前整体贪心解要比非贪心解要好
- 时间复杂度：\(O(n)\)
代码
// Problem: E1. Permutation Minimization by Deque
// Contest: Codeforces - Codeforces Round #744 (Div. 3)
// URL: https://codeforces.com/contest/1579/problem/E1
// Memory Limit: 256 MB
// Time Limit: 2000 ms
//
// Powered by CP Editor (https://cpeditor.org)
// %%%Skyqwq
#include <bits/stdc++.h>
//#define int long long
#define help {cin.tie(NULL); cout.tie(NULL);}
#define pb push_back
#define fi first
#define se second
#define mkp make_pair
using namespace std;
typedef long long LL;
typedef pair<int, int> PII;
typedef pair<LL, LL> PLL;
template <typename T> bool chkMax(T &x, T y) { return (y > x) ? x = y, 1 : 0; }
template <typename T> bool chkMin(T &x, T y) { return (y < x) ? x = y, 1 : 0; }
template <typename T> void inline read(T &x) {
int f = 1; x = 0; char s = getchar();
while (s < '0' || s > '9') { if (s == '-') f = -1; s = getchar(); }
while (s <= '9' && s >= '0') x = x * 10 + (s ^ 48), s = getchar();
x *= f;
}
int main()
{
int t;
for(cin>>t;t;t--)
{
int n;
cin>>n;
deque<int> q;
while(n--)
{
int x;
cin>>x;
if(q.size()&&x<q.front())q.push_front(x);
else
q.pb(x);
}
while(q.size())cout<<q.front()<<' ',q.pop_front();
puts("");
}
return 0;
}
E2. Array Optimization by Deque
输入 \(n\) 个数，按照输入顺序根据双端队列的插入规则使得逆序对最小
解题思路
贪心
遍历每个数，而前面的数已经固定，当前数要么放前面要么放后面，考虑当前数对答案的贡献，由于当前数与之前的数组成的整体对后面数的逆序对没有影响，所以当前数对答案的贡献越小越好，即选择放前面和放后面逆序对较小的位置
- 时间复杂度：\(O(nlogn)\)
代码
// Problem: E2. Array Optimization by Deque
// Contest: Codeforces - Codeforces Round #744 (Div. 3)
// URL: https://codeforces.com/contest/1579/problem/E2
// Memory Limit: 256 MB
// Time Limit: 2000 ms
//
// Powered by CP Editor (https://cpeditor.org)
// %%%Skyqwq
#include <bits/stdc++.h>
//#define int long long
#define help {cin.tie(NULL); cout.tie(NULL);}
#define pb push_back
#define fi first
#define se second
#define mkp make_pair
using namespace std;
typedef long long LL;
typedef pair<int, int> PII;
typedef pair<LL, LL> PLL;
template <typename T> bool chkMax(T &x, T y) { return (y > x) ? x = y, 1 : 0; }
template <typename T> bool chkMin(T &x, T y) { return (y < x) ? x = y, 1 : 0; }
template <typename T> void inline read(T &x) {
int f = 1; x = 0; char s = getchar();
while (s < '0' || s > '9') { if (s == '-') f = -1; s = getchar(); }
while (s <= '9' && s >= '0') x = x * 10 + (s ^ 48), s = getchar();
x *= f;
}
const int N=2e5+5;
int t,n,a[N],tr[N];
vector<int> xs;
int find(int x)
{
return lower_bound(xs.begin(),xs.end(),x)-xs.begin()+1;
}
int ask(int x)
{
int res=0;
for(;x;x-=x&-x)res+=tr[x];
return res;
}
void add(int x,int y)
{
for(;x<=n;x+=x&-x)tr[x]+=y;
}
int main()
{
help;
for(cin>>t;t;t--)
{
cin>>n;
xs.clear();
for(int i=1;i<=n;i++)
{
cin>>a[i];
xs.pb(a[i]);
tr[i]=0;
}
sort(xs.begin(),xs.end());
xs.erase(unique(xs.begin(),xs.end()),xs.end());
for(int i=1;i<=n;i++)a[i]=find(a[i]);
LL res=0;
for(int i=1;i<=n;i++)
{
int x=ask(a[i]-1),y=i-1-ask(a[i]);
res+=min(x,y);
add(a[i],1);
}
cout<<res<<'\n';
}
return 0;
}
