树状数组求LIS模板
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如果数组元素较大,需要离散化。
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <vector>
#include <iomanip>
#define ALL(x) (x).begin(), (x).end()
#define rt return
#define dll(x) scanf("%I64d",&x)
#define xll(x) printf("%I64d\n",x)
#define sz(a) int(a.size())
#define all(a) a.begin(), a.end()
#define rep(i,x,n) for(int i=x;i<n;i++)
#define repd(i,x,n) for(int i=x;i<=n;i++)
#define pii pair<int,int>
#define pll pair<long long ,long long>
#define gbtb ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
#define MS0(X) memset((X), 0, sizeof((X)))
#define MSC0(X) memset((X), '\0', sizeof((X)))
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define eps 1e-6
#define gg(x) getInt(&x)
#define chu(x) cout<<"["<<#x<<" "<<(x)<<"]"<<endl
using namespace std;
typedef long long ll;
ll gcd(ll a,ll b)return b?gcd(b,a%b):a;
ll lcm(ll a,ll b)return a/gcd(a,b)*b;
ll powmod(ll a,ll b,ll MOD)ll ans=1;while(b)if(b%2)ans=ans*a%MOD;a=a*a%MOD;b/=2;return ans;
inline void getInt(int* p);
const int maxn=1000010;
const int inf=0x3f3f3f3f;
/*** TEMPLATE CODE * * STARTS HERE ***/
ll tree[maxn];
int n;
ll a[maxn];
ll lowbit(ll x)
return x&(-x);
ll query(ll x)
ll res=0ll;
while(x)
res=max(res,tree[x]);
x-=lowbit(x);
return res;
void modify(ll id,ll x)
while(id<maxn)
tree[id]=max(tree[id],x);
id+=lowbit(id);
int main()
//freopen("D:\\common_text\\code_stream\\in.txt","r",stdin);
//freopen("D:\\common_text\code_stream\\out.txt","w",stdout);
gbtb;
cin>>n;
ll ans=0ll;
ll x;
repd(i,1,n)
cin>>a[i];
x=query(a[i])+1;
ans=max(ans,x);
modify(a[i],x);
cout<<ans<<endl;
return 0;
inline void getInt(int* p)
char ch;
do
ch = getchar();
while (ch == ' ' || ch == '\n');
if (ch == '-')
*p = -(getchar() - '0');
while ((ch = getchar()) >= '0' && ch <= '9')
*p = *p * 10 - ch + '0';
else
*p = ch - '0';
while ((ch = getchar()) >= '0' && ch <= '9')
*p = *p * 10 + ch - '0';
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