E. Two Teams
time limit per test
2 seconds
memory limit per test
256 megabytes
There are nn students standing in a row. Two coaches are forming two teams — the first coach chooses the first team and the second coach chooses the second team.
The ii-th student has integer programming skill aiai. All programming skills are distinct and between 11 and nn, inclusive.
Firstly, the first coach will choose the student with maximum programming skill among all students not taken into any team, and kk closest students to the left of him and kk closest students to the right of him (if there are less than kk students to the left or to the right, all of them will be chosen). All students that are chosen leave the row and join the first team. Secondly, the second coach will make the same move (but all students chosen by him join the second team). Then again the first coach will make such move, and so on. This repeats until the row becomes empty (i. e. the process ends when each student becomes to some team).
Your problem is to determine which students will be taken into the first team and which students will be taken into the second team.
Input
The first line of the input contains two integers nn and kk (1≤k≤n≤2⋅1051≤k≤n≤2⋅105) — the number of students and the value determining the range of chosen students during each move, respectively.
The second line of the input contains nn integers a1,a2,…,ana1,a2,…,an (1≤ai≤n1≤ai≤n), where aiai is the programming skill of the ii-th student. It is guaranteed that all programming skills are distinct.
Output
Print a string of nn characters; ii-th character should be 1 if ii-th student joins the first team, or 2 otherwise.
Examples
input
5 2 2 4 5 3 1
output
11111
input
5 1 2 1 3 5 4
output
22111
input
7 1 7 2 1 3 5 4 6
output
1121122
input
5 1 2 4 5 3 1
output
21112
Note
In the first example the first coach chooses the student on a position 33, and the row becomes empty (all students join the first team).
In the second example the first coach chooses the student on position 44, and the row becomes [2,1][2,1] (students with programming skills [3,4,5][3,4,5] join the first team). Then the second coach chooses the student on position 11, and the row becomes empty (and students with programming skills [1,2][1,2] join the second team).
In the third example the first coach chooses the student on position 11, and the row becomes [1,3,5,4,6][1,3,5,4,6] (students with programming skills [2,7][2,7] join the first team). Then the second coach chooses the student on position 55, and the row becomes [1,3,5][1,3,5] (students with programming skills [4,6][4,6] join the second team). Then the first coach chooses the student on position 33, and the row becomes [1][1] (students with programming skills [3,5][3,5] join the first team). And then the second coach chooses the remaining student (and the student with programming skill 11 joins the second team).
In the fourth example the first coach chooses the student on position 33, and the row becomes [2,1][2,1] (students with programming skills [3,4,5][3,4,5] join the first team). Then the second coach chooses the student on position 11, and the row becomes empty (and students with programming skills [1,2][1,2] join the second team).
用链表
题意：轮到一个教练选人的时候，先定位到当前剩余人中 programming skills 的值最大的那个，再加上其左右两边k个人。
AC代码：
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <vector>
#include <string>
#include <cstring>
#include <stack>
#include <map>
#include <set>
#include <cassert>
using namespace std;
#define rep(i,a,n) for (int i=a;i<n;i++)
#define per(i,a,n) for (int i=n-1;i>=a;i--)
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
typedef vector<int> VI;
typedef long long ll;
typedef pair<int, int> PII;
typedef pair<long, long> PLL;
const ll mod = 1e9 + 7;
const int N = 1e6;
ll n, m, k;
int ans[200006];
struct node {
int id;
int a;
};
bool cmp(const node a, const node b) {
return a.a < b.a;
}
vector<node> v;
int Next[200006];
int pre[200006];
int main() {
cin >> n >> k;
int key;
memset(ans, -1, sizeof(ans));
for (int i = 1; i <= n; i++) {
cin >> key;
v.pb(node{ i, key });
}
sort(v.begin(), v.end(), cmp);
Next[0] = 1;
for (int i = 1; i<n; i++) {
pre[i] = i - 1;
Next[i] = i + 1;
}
pre[n] = n - 1;
Next[n] = 0;
int flag = 1;
while (1) {
if (!Next[0]) break;
int Max = 0, Maxi = 0;
for (int i = v.size() - 1; i >= 0; i--) {
if (ans[v[i].id] == -1) {
Maxi = v[i].id;
v.pop_back();
break;
}
else
v.pop_back();
}
int tot = 0;
for (int i = Next[Maxi]; i != 0 && tot<k; i = Next[i]) {
ans[i] = flag;
Next[Maxi] = Next[i];
pre[Next[i]] = Maxi;
tot++;
}
tot = 0;
for (int i = pre[Maxi]; i != 0 && tot<k; i = pre[i]) {
ans[i] = flag;
pre[Maxi] = pre[i];
Next[pre[i]] = Maxi;
tot++;
}
ans[Maxi] = flag;
Next[pre[Maxi]] = Next[Maxi];
pre[Next[Maxi]] = pre[Maxi];
/*for (int i = Next[0]; i != 0; i = Next[i]) {
cout << i<<' ';
}
cout << '\n';
for (int i = 1; i <= n; i++) {
cout << ans[i] << ' ';
}
cout << '\n';*/
flag ^= 1;
}
for (int i = 1; i <= n; i++) {
if (ans[i]) cout << 1;
else cout << 2;
}
return 0;
}
