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
", q.back().c_str()); q.pop_back(); }
  else while(!q.empty()){  printf("%s
", q.front().c_str()); q.pop_front(); }
  while(!st.empty()){ printf("%s
", st.top().c_str());  st.pop(); }
  for(int i = 0; i < v.sz; ++i)  printf("%s
", v[i].c_str());
  return 0;
}

  

原文地址:https://www.cnblogs.com/dwtfukgv/p/8687711.html