【DP】 HDOJ 3530 Subsequence

维护两个单调队列就行了。。

#include <iostream>  
#include <queue>  
#include <stack>  
#include <map>  
#include <set>  
#include <bitset>  
#include <cstdio>  
#include <algorithm>  
#include <cstring>  
#include <climits>  
#include <cstdlib>
#include <cmath>
#include <time.h>
#define maxn 100005
#define maxm 400005
#define eps 1e-10
#define mod 1000000007
#define INF 999999999
#define lowbit(x) (x&(-x))
#define mp mark_pair
#define ls o<<1
#define rs o<<1 | 1
#define lson o<<1, L, mid  
#define rson o<<1 | 1, mid+1, R  
//typedef vector<int>::iterator IT;
typedef long long LL;
//typedef int LL;
using namespace std;
LL qpow(LL a, LL b){LL res=1,base=a;while(b){if(b%2)res=res*base;base=base*base;b/=2;}return res;}
LL powmod(LL a, LL b){LL res=1,base=a;while(b){if(b%2)res=res*base%mod;base=base*base%mod;b/=2;}return res;}
void scanf(int &__x){__x=0;char __ch=getchar();while(__ch==' '||__ch=='\n')__ch=getchar();while(__ch>='0'&&__ch<='9')__x=__x*10+__ch-'0',__ch = getchar();}
LL gcd(LL _a, LL _b){if(!_b) return _a;else return gcd(_b, _a%_b);}
// head

int q1[maxn], q2[maxn], num[maxn];
int n, m, k;
void read(void)
{
	for(int i = 1; i <= n; i++) scanf("%d", &num[i]);
}
void work(void)
{
	int t1, t2, l1, l2, r1, r2, mx;
	t1 = t2 = l1 = l2 = r1 = r2 = mx = 0;
	for(int i = 1; i <= n; i++) {
		while(l1 < t1 && num[q1[t1-1]] < num[i]) t1--;
		q1[t1++] = i;
		while(l2 < t2 && num[q2[t2-1]] > num[i]) t2--;
		q2[t2++] = i;
		while(num[q1[l1]] - num[q2[l2]] > k) {
			if(q1[l1] < q2[l2]) r1 = q1[l1++];
			else r2 = q2[l2++];
		}
		if(num[q1[l1]] - num[q2[l2]] >= m) mx = max(mx, i - max(r1, r2));
	}
	printf("%d\n", mx);
}
int main(void)
{
	while(scanf("%d%d%d", &n, &m, &k)!=EOF) {
		read();
		work();
	}
	return 0;
}