Codeforces 1154E Two Teams

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

题目大意：

　　有n个队员，编号1~n，每个人的能力各自对应1~n中的一个数，每个人的能力都不相同。有1号教练和2号教练，他们轮流从剩余队伍里选人，轮到某位教练选时，它总是选剩余队员中能力最强的人和他左右各k个人。问选完的时候每个人的组号。

分析：

　　什么结构能快速查找到队员呢？当然是数组啦！什么结构能频繁删改呢？当然是链表啦！所以就用承载在数组上的链表来做啦！

代码如下：

#pragma GCC optimize("Ofast")
#include <bits/stdc++.h>
using namespace std;
 
#define INIT() std::ios::sync_with_stdio(false);std::cin.tie(0);
#define Rep(i,n) for (int i = 0; i < (n); ++i)
#define For(i,s,t) for (int i = (s); i <= (t); ++i)
#define rFor(i,t,s) for (int i = (t); i >= (s); --i)
#define ForLL(i, s, t) for (LL i = LL(s); i <= LL(t); ++i)
#define rForLL(i, t, s) for (LL i = LL(t); i >= LL(s); --i)
#define foreach(i,c) for (__typeof(c.begin()) i = c.begin(); i != c.end(); ++i)
#define rforeach(i,c) for (__typeof(c.rbegin()) i = c.rbegin(); i != c.rend(); ++i)
 
#define pr(x) cout << #x << " = " << x << "  "
#define prln(x) cout << #x << " = " << x << endl
 
#define LOWBIT(x) ((x)&(-x))
 
#define ALL(x) x.begin(),x.end()
#define INS(x) inserter(x,x.begin())
 
#define ms0(a) memset(a,0,sizeof(a))
#define msI(a) memset(a,inf,sizeof(a))
#define msM(a) memset(a,-1,sizeof(a))

#define MP make_pair
#define PB push_back
#define ft first
#define sd second
 
template<typename T1, typename T2>
istream &operator>>(istream &in, pair<T1, T2> &p) {
    in >> p.first >> p.second;
    return in;
}
 
template<typename T>
istream &operator>>(istream &in, vector<T> &v) {
    for (auto &x: v)
        in >> x;
    return in;
}
 
template<typename T1, typename T2>
ostream &operator<<(ostream &out, const std::pair<T1, T2> &p) {
    out << "[" << p.first << ", " << p.second << "]" << "
";
    return out;
}
 
typedef long long LL;
typedef unsigned long long uLL;
typedef pair< double, double > PDD;
typedef set< int > SI;
typedef vector< int > VI;
const double EPS = 1e-10;
const int inf = 1e9 + 9;
const LL mod = 1e9 + 7;
const int maxN = 2e5 + 7;
const LL ONE = 1;

struct Node{
    int value, pos;
    Node* prev = NULL;
    Node* next = NULL;
};

int n, k, ans[maxN];
Node nodes[maxN];
int to[maxN]; 
int f = 1;

int main(){
    scanf("%d%d
", &n, &k);
    For(i, 1, n) {
        scanf("%d", &nodes[i].value);
        nodes[i].pos = i;
        to[nodes[i].value] = i;
    }
    
    Rep(i, n + 1) nodes[i].next = &nodes[i + 1];
    Rep(i, n + 1) nodes[i + 1].prev = &nodes[i];
    
    int i = n;
    while(i >= 1) {
        if(ans[to[i]] != 0) {
            --i;
            continue;
        }
        
        ans[to[i]] = f;
        // 处理左边k个 
        int cnt = 0;
        Node *p = nodes[to[i]].prev;
        while(p->prev != NULL && cnt < k) {
            ans[p->pos] = f;
            p = p->prev;
            ++cnt;
        }
        // 处理右边k个 
        cnt = 0;
        Node *q = nodes[to[i]].next;
        while(q->next != NULL && cnt < k) {
            ans[q->pos] = f;
            q = q->next;
            ++cnt;
        }
        
        // 删掉选走的 
        p->next = q;
        q->prev = p;
        
        f = f == 1 ? 2 : 1;
    }
    
    For(i, 1, n) printf("%d", ans[i]);
    printf("
");
    return 0;
}