大意: 给定序列, 求选出一个最长的子序列, 使得任选两个[1,8]的数字, 在子序列中的出现次数差不超过1, 且子序列中相同数字连续.

 

 

正解是状压dp, 先二分转为判断[1,8]出现次数>=x是否成立, 再dp求出前i位匹配状态S长度为x+1的数字个数的最大值, 特判一下最低次数为0的情况. 这题打了好久, 太菜了.......

#include <iostream>
#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'
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
typedef bitset<10> btc;
const int P = 1e9+7, INF = 0xbcbcbcbc;
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;}
//head



const int N = 1e3+10, S = (1<<8)-1;
int n, ans;
int a[N], dp[N][S+1], dig[S+1], f[N][N][9], g[S+1];
void chkmax(int &x, int y) {x=max(x,y);}
void chkmin(int &x, int y) {x=min(x,y);}
int chk(int x) {
	memset(dp, 0xbc, sizeof dp);
	dp[0][0] = 0;
	REP(i,1,n) REP(j,0,S-1) if (dp[i-1][j]!=INF) {
		for (int k=~j&S, t; k; k^=t) {
			t = k&-k;
			int p1 = f[i][x][dig[t]], p2 = f[i][x+1][dig[t]];
			if (p1<=n) chkmax(dp[p1][j^t], dp[i-1][j]);
			if (p2<=n) chkmax(dp[p2][j^t], dp[i-1][j]+1);
		}
	}
	int r = INF;
	REP(i,1,n) chkmax(r,dp[i][S]);
	ans = max(ans, r+8*x);
	return r!=INF;
}


int main() {
	REP(i,0,7) dig[1<<i]=i+1;
	scanf("%d", &n);
	REP(i,1,n) scanf("%d", a+i);
	memset(f,0x3f,sizeof f);
	PER(i,1,n) REP(j,1,n) REP(k,1,8) {
		if (a[i]==k) f[i][1][k]=i,f[i][j][k]=f[i+1][j-1][k];
		else chkmin(f[i][j][k],f[i+1][j][k]);
	}
	g[0] = 1;
	REP(i,1,n) REP(j,0,S) if (!(j>>a[i]-1&1)) {
		g[j^1<<a[i]-1] |= g[j];
	}
	REP(i,0,S) if (g[i]) ans=max(ans,__builtin_popcount(i));
	int l=1, r=n/8;
	while (l<=r) {
		if (chk(mid)) l=mid+1;
		else r=mid-1;
	}
	printf("%d\n", ans);
}