牛客 158F 青蛙 (贪心)

显然存在一个最优解满足所有青蛙在连续的一段, 每次由最左侧青蛙跳向下一格.

然后二分或者双指针即可求出答案.

#include <iostream>
#include <sstream>
#include <algorithm>
#include <cstdio>
#include <math.h>
#include <set>
#include <map>
#include <queue>
#include <string>
#include <string.h>
#include <bitset>
#define REP(i,a,n) for(int i=a;i<=n;++i)
#define PER(i,a,n) for(int i=n;i>=a;--i)
#define hr putchar(10)
#define pb push_back
#define lc (o<<1)
#define rc (lc|1)
#define mid ((l+r)>>1)
#define ls lc,l,mid
#define rs rc,mid+1,r
#define x first
#define y second
#define io std::ios::sync_with_stdio(false)
#define endl '\n'
#define DB(a) ({REP(__i,1,n) cout<<a[__i]<<' ';hr;})
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int P = 1e9+7, P2 = 998244353, INF = 0x3f3f3f3f;
ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}
ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;}
ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;}
inline int rd() {int x=0;char p=getchar();while(p<'0'||p>'9')p=getchar();while(p>='0'&&p<='9')x=x*10+p-'0',p=getchar();return x;}
//head

const int N = 1e6+10;
int a[N], f[N];

int main() {
	int t;
	scanf("%d", &t);
	while (t--) {
		int n, m, d;
		scanf("%d%d%d", &n, &m, &d),m+=2;
		REP(i,1,m) scanf("%d", a+i);
		REP(i,1,m) {
			if (a[i]-a[1]<=d) f[i] = 1;
			else {
				int j = f[i-1];
				while (a[i]-a[j]>d) ++j;
				f[i] = j+1;
			}
		}
		printf("%d\n", m-f[m]+1);
	}
}