POJ 2823 Sliding Window (单调队列)

题意：给定一个序列，从左到右每次的滑动一个窗口，最大值和最小值是多少。

析：普通的方法可能会超时，维护两个单调队列，一个单调递增的，一个单调递减，每次把最值保存下来。

也可以用线段树，RMQ等数据结构，每次查询区间的最小值和最大值。POJ 交G++ 死活超时，交C++才过。

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>
#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 freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;

typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 1e6 + 10;
const int mod = 1e9 + 7;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -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;
}

int a[maxn];
int q1[maxn], q2[maxn];
int ans1[maxn], ans2[maxn];

int main(){
  while(scanf("%d %d", &n, &m) == 2){
    int fro1 = 1, rear1 = 0;
    int fro2 = 1, rear2 = 0;
    for(int i = 0; i < m; ++i){
      scanf("%d", a+i);
      while(fro1 <= rear1 && a[i] <= a[q1[rear1]]) --rear1;
      while(fro2 <= rear2 && a[i] >= a[q2[rear2]]) --rear2;
      q1[++rear1] = i;
      q2[++rear2] = i;
    }
    ans1[0] = a[q1[fro1]];
    ans2[0] = a[q2[fro2]];
    for(int i = m; i < n; ++i){
      scanf("%d", a+i);
      while(fro1 <= rear1 && a[i] <= a[q1[rear1]]) --rear1;
      while(fro2 <= rear2 && a[i] >= a[q2[rear2]]) --rear2;
      q1[++rear1] = q2[++rear2] = i;
      while(q1[fro1] <= i-m) ++fro1;
      while(q2[fro2] <= i-m) ++fro2;
      ans1[i-m+1] = a[q1[fro1]];
      ans2[i-m+1] = a[q2[fro2]];
    }
    for(int i = 0; i <= n-m; ++i){
      if(i)  putchar(' ');
      printf("%d", ans1[i]);
    }
    printf("\n");
    for(int i = 0; i <= n-m; ++i){
      if(i)  putchar(' ');
      printf("%d", ans2[i]);
    }
    printf("\n");
  }
  return 0;
}
/*
10 3
154 5 54545154 51 545 1545464 54 2 54 54
*/