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 a[j]-a[i]>=j-i a[j]−a[i]>=j−i,化简得 a [ j ] − j > a [ i ] − i a[j]-j>a[i]-i 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 b[i]=a[i]-i 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;
}