SGU 271 Book Pile (双端队列)

题意：n，m，k，表示有一个长度为 n 的序列，有 m 个操作，操作有 2 种，第一种是 ADD 在前面添加一个串，第二种是把前 k 个进行翻转，问你最后的序列是什么样的。

析：很明显，如果直接模拟，肯定会超时，由于 k 是固定的，我们就可以前 k 个串，如果没有翻转，那么就把添加的串方法直接放到双端队列前面，然后把双端队列后面那个串再拿出去，如果翻转了，就把添加的串方法直接放到双端队列后面，然后把双端队列前面那个串再拿出去。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")

#include <cstdio>

#include <string>

#include <cstdlib>

#include <cmath>

#include <iostream>

#include <cstring>

#include <set>

#include <queue>

#include <algorithm>

#include <vector>

#include <map>

#include <cctype>

#include <cmath>

#include <stack>

#include <sstream>

#include <list>

#include <assert.h>

#include <bitset>

#include <numeric>

#define debug() puts("++++")

#define gcd(a, b) __gcd(a, b)

#define lson l,m,rt<<1

#define rson m+1,r,rt<<1|1

#define fi first

#define se second

#define pb push_back

#define sqr(x) ((x)*(x))

#define ms(a,b) memset(a, b, sizeof a)

#define sz size()

#define be begin()

#define ed end()

#define pu push_up

#define pd push_down

#define cl clear()

#define lowbit(x) -x&x

//#define all 1,n,1

#define FOR(i,n,x)  for(int i = (x); i < (n); ++i)

#define freopenr freopen("in.in", "r", stdin)

#define freopenw freopen("out.out", "w", stdout)

using namespace std;

typedef long long LL;

typedef unsigned long long ULL;

typedef pair<int, int> P;

const int INF = 0x3f3f3f3f;

const LL LNF = 1e17;

const double inf = 1e20;

const double PI = acos(-1.0);

const double eps = 1e-8;

const int maxn = 1e5 + 10;

const int maxm = 1e6 + 10;

const LL mod = 1000000007;

const int dr[] = {-1, 1, 0, 0, 1, 1, -1, -1};

const int dc[] = {0, 0, 1, -1, 1, -1, 1, -1};

const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};

int n, m;

const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};

const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};

inline bool is_in(int r, int c) {

  return r >= 0 && r < n && c >= 0 && c < m;

}

inline int readInt(){ int x;  scanf("%d", &x);  return x; }

vector<string> v;

stack<string> st;

deque<string> q;

int main(){

  int k;

  scanf("%d %d %d", &n, &m, &k);

  char s[30];

  int cnt = 0;

  for(int i = 0; i < n; ++i){

    scanf("%s", s);

    if(q.sz < k)  q.push_back(s);

    else v.pb(s);

  }

  char t[30];

  while(m--){

    scanf("%s", t);

    if(t[0] == 'R')  cnt ^= 1;

    else {

      t[strlen(t)-1] = 0;

      char *s = t + 4;

      if(cnt == 0){

        q.push_front(s);

        if(q.sz > k){

          st.push(q.back());  q.pop_back();

        }

      }

      else{

        q.push_back(s);

        if(q.sz > k){

          st.push(q.front());  q.pop_front();

        }

      }

    }

  }

  if(cnt)  while(!q.empty()){  printf("%s\n", q.back().c_str()); q.pop_back(); }

  else while(!q.empty()){  printf("%s\n", q.front().c_str()); q.pop_front(); }

  while(!st.empty()){ printf("%s\n", st.top().c_str());  st.pop(); }

  for(int i = 0; i < v.sz; ++i)  printf("%s\n", v[i].c_str());

  return 0;

}