Codeforces 552 E. Two Teams

最新推荐文章于 2025-09-12 17:31:31 发布

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最新推荐文章于 2025-09-12 17:31:31 发布

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原文链接：http://www.cnblogs.com/kongbb/p/10743056.html

本文解析了CodeForces竞赛中一道关于选择学生队伍的算法题。通过使用类似双链表的数据结构，实现了快速查找并标记学生的过程，最终确定每个学生所属的队伍。

There are

The

Firstly, the first coach will choose the student with maximum programming skill among all students not taken into any team, and

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

The second line of the input contains

Output

Print a string of

Examples

input

5 2
2 4 5 3 1

output

input

5 1
2 1 3 5 4

output

input

7 1
7 2 1 3 5 4 6

output

input

5 1
2 4 5 3 1

output

Note

In the first example the first coach chooses the student on a position

In the second example the first coach chooses the student on position

In the third example the first coach chooses the student on position

In the fourth example the first coach chooses the student on position

题目链接： http://codeforces.com/contest/1154/problem/E

题意：

　　给你n个值（1~n且不重复），选中最大的数，并将它左边k个元素和右边k个元素（可能不足，不足的部分不考虑）选中标记为 1 ，这2*k+1个元素同时从数组中删除。重复操作，但是标记为2，结束一轮。

　　重复任意多轮，直至所有数都被标记(1 or 2)。

当时并没有好的思路，赛后看的别人的代码才会的。

大佬的AK日常： https://www.jianshu.com/p/fd764cf686e9

考虑到数据是1~n的全排列，所以我们可以用数组来将元素的位置离散出来。

实现类似于双链表的数据结构，用于快速找到当前元素的左右相邻元素。

#include<cstdio>
#include<cstring>
#include<iostream>
using namespace std;

const int maxn = 2e5+5;
int n, k;
int sk[maxn], le[maxn], ri[maxn];
int arr[maxn], ans[maxn];

void remo(int ind){
    le[ri[ind]] = le[ind];
    ri[le[ind]] = ri[ind];
}
int main(){
    memset(ans,0,sizeof(ans));
    scanf("%d%d",&n,&k);
    for(int i=1;i<=n;i++){
        scanf("%d",&arr[i]);
        sk[arr[i]] = i;
        le[i] = i-1;
        ri[i] = i+1;
    }
    ans[0] = ans[n+1] = -1;// 边界
    int team = 1;
    for(int i=n;i>=0;i--){
        int index = sk[i];
        if(ans[index])
            continue;
        remo(index);
        ans[index] = team;
        int l = le[index], r = ri[index];
        for(int i=0;i<k;i++,l=le[l]){
            if(ans[l])
                break;
            remo(l);
            ans[l] = team;
        }
        for(int i=0;i<k;i++,r=ri[r]){
            if(ans[r])
                break;
                remo(r);
                ans[r] = team;
        }
        team = 3 - team;
    }
    for(int i=1;i<=n;i++)
        printf("%d",ans[i]);
    putchar('\n');
    return 0;
}