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 << "]" << "\n";
     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-;
 const int inf = 1e9 + ;
 const LL mod = 1e9 + ;
 const int maxN = 2e5 + ;
 const LL ONE = ;

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

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

 int main(){
     scanf("%d%d\n", &n, &k);
     For(i, , n) {
         scanf("%d", &nodes[i].value);
         nodes[i].pos = i;
         to[nodes[i].value] = i;
     }

     Rep(i, n + ) nodes[i].next = &nodes[i + ];
     Rep(i, n + ) nodes[i + ].prev = &nodes[i];

     int i = n;
     while(i >= ) {
         if(ans[to[i]] != ) {
             --i;
             continue;
         }

         ans[to[i]] = f;
         // 处理左边k个
         int cnt = ;
         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 = ;
         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 ==  ?  : ;
     }

     For(i, , n) printf("%d", ans[i]);
     printf("\n");
     return ;
 }