Two Merged Sequences CodeForces - 1144G (暴力)

大意: 给定序列, 求划分为一个严格递增子序列和一个严格递减子序列, 可以为空.

 

跟 125D 类似的一个题, 直接暴力dfs, 用当前序列长度来剪枝, 状态不会太多, 但是会被一些数据卡掉, 特判一下小数据时不剪枝.

#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, 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



#ifdef ONLINE_JUDGE
const int N = 1e6+10;
#else
const int N = 111;
#endif

int n, a[N], ans[N];
vector<int> x, y;
map<int,bool> vx[N], vy[N];
int dfs(int d) {
	if (d>n) return 1;
	if (x.empty()||a[x.back()]<a[d]) {
		x.pb(d);
		if (!vx[d].count(int(x.size()))) {
			vx[d][int(x.size())]=1;
			if (dfs(d+1)) return 1;
		}
		else if (d<=10) {
			if (dfs(d+1)) return 1;
		}
		x.pop_back();
	}
	if (y.empty()||a[y.back()]>a[d]) {
		y.pb(d);
		if (!vy[d].count(int(y.size()))) {
			vy[d][int(y.size())]=1;
			if (dfs(d+1)) return 1;
		}
		else if (d<=10) {
			if (dfs(d+1)) return 1;
		}
		y.pop_back();
	}
	return 0;
}

int main() {
	scanf("%d", &n);
	REP(i,1,n) scanf("%d", a+i);
	if (dfs(1)) {
		for (auto &&t:y) ans[t]=1;
		puts("YES");
		REP(i,1,n) printf("%d ", ans[i]);hr;
	}
	else puts("NO");
}