HDU 5356 序列变换

思路

这个题是最长上升子序列的变形版本，要保证严格递增，需要满足
$a[j]-a[i]>=j-i$,化简得$a[j]-j>a[i]-i$,所以我们只需要求得序列b的LIS即可($b[i]=a[i]-i$)，不过你要是用动态规划写的话会T掉，求LIS需要用到贪心+二分的解法。如果不了解的话，可以看下这篇最长上升子序列的两种复杂度解法

代码

• 使用upper_bound()的版本

#pragma GCC optimize(2)
#include<bits/stdc++.h>

using namespace std;

typedef long long ll;
typedef unsigned long ul;
typedef unsigned long long ull;
#define pi acos(-1.0)
#define e exp(1.0)
#define pb push_back
#define mk make_pair
#define fir first
#define sec second
#define scf scanf
#define prf printf
typedef pair<ll,ll> pa;
const ll INF=0x3f3f3f3f3f3f3f3f;
const ll MAX_T=12;
const ll MAX_N=1e5+7;
ll T,N;
ll A[MAX_N],B[MAX_N];
ll do_LIS(){
	ll i,j,k,cnt=0;
	for(i=0;i<N;i++){
		ll pos=upper_bound(B,B+cnt,A[i])-B;
		if(pos==cnt){
			cnt++;
		}
		B[pos]=A[i];
	}
	return cnt;
}
int main()
{
//  freopen(".../.txt","w",stdout);
//  freopen(".../.txt","r",stdin);
//	ios::sync_with_stdio(false);
	ll i,j,k=0;
	scf("%lld",&T);
	while(T--){
		scf("%lld",&N);
		for(i=0;i<N;i++){
			scf("%lld",&A[i]);
			A[i]-=i;
		}
		ll res=N-do_LIS();
		prf("Case #%lld:\n",++k);
		prf("%lld\n",res);
	}
	return 0;
}

• 这个是手写二分的版本

#pragma GCC optimize(2)
#include<bits/stdc++.h>

using namespace std;

typedef long long ll;
typedef unsigned long ul;
typedef unsigned long long ull;
#define pi acos(-1.0)
#define e exp(1.0)
#define pb push_back
#define mk make_pair
#define fir first
#define sec second
#define scf scanf
#define prf printf
typedef pair<ll,ll> pa;
const ll INF=0x3f3f3f3f3f3f3f3f;
const ll MAX_T=12;
const ll MAX_N=1e5+7;
ll T,N;
ll A[MAX_N],B[MAX_N];
ll do_LIS(){
	ll i,j,k,cnt=0;
	for(i=0;i<N;i++){
		ll L=0,R=cnt-1,mid,res=-1;
		while(L<=R){
			mid=L+(R-L)/2;
			if(B[mid]>A[i]){
				res=mid;
				R=mid-1;
			}
			else{
				L=mid+1;
			}
		}
		if(res==-1){
			B[cnt]=A[i];
			cnt++;
		}
		else
		B[res]=A[i];
	}
	return cnt;
}
int main()
{
//  freopen(".../.txt","w",stdout);
//  freopen(".../.txt","r",stdin);
//	ios::sync_with_stdio(false);
	ll i,j,k=0;
	scf("%lld",&T);
	while(T--){
		scf("%lld",&N);
		for(i=0;i<N;i++){
			scf("%lld",&A[i]);
			A[i]-=i;
		}
		ll res=N-do_LIS();
		prf("Case #%lld:\n",++k);
		prf("%lld\n",res);
	}
	return 0;
}