HDU 3706 Second My Problem First (单调队列)

题意：求给定的一个序列中最长子序列，该子序列的最大值和最小值介于m和k之间。

析：用两个单调队列来维护一个最小值，一个最大值，然后每次更新即可。

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-5;
const int maxn = 100000 + 10;
const int mod = 1e6;
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 q1[maxn], q2[maxn];
int a[maxn], k;

int main(){
  while(scanf("%d %d %d", &n, &m, &k) == 3){
    int ans = 0;
    int fro1 = 0;  int rear1 = 0;
    int fro2 = 0;  int rear2 = 0;
    int tmp = 1;
    for(int i = 1; i <= n; ++i){
      scanf("%d", a+i);
      while(fro1 < rear1 && a[q1[rear1-1]] > a[i])  --rear1;  //increase
      while(fro2 < rear2 && a[q2[rear2-1]] < a[i])  --rear2;  //decrease
      q1[rear1++] = q2[rear2++] = i;
      while(fro1 < rear1 && fro2 < rear2 && a[q2[fro2]]-a[q1[fro1]] > k){
        if(q1[fro1] < q2[fro2])  tmp = q1[fro1++] + 1;
        else tmp = q2[fro2++] + 1;
      }
      if(fro1 < rear1 && fro2 < rear2 && a[q2[fro2]]-a[q1[fro1]] >= m)  ans = max(ans, i - tmp + 1);
    }
    printf("%d\n", ans);
  }
  return 0;
}