CF384 div2 E. Vladik and cards

题意

给你一个$1-8$的排列，求一个满足条件的最长子序列
每种数字的差小于等于$1$，并且每种数字之内是连续的

解法

首先单纯认为用$dp$肯定不行的
所以应该考虑二分答案（所求长度具有二分性）
再用dp判断是否可行，这个dp很简单就是$dp[N][1<<8]$

#include<iostream>
#include<cstdio>
#include<cstring>
#include<vector>
using namespace std;
const int N = 1005;
const int INF = 0x3f3f3f3f;

int n;
vector<int> in[10];
int dp[N][300];
int a[N];
int cur[10];
void gmax(int &a, int b) {
    if(a < b) a = b;
}
int ok(int len) {
    for(int i = 0; i < 10; ++i) cur[i] = 0;
    memset(dp,-1,sizeof(dp));
    dp[0][0] = 0;
    for(int i = 0; i < n; ++i) {
        for(int j = 0; j < 256; ++j) {
            if(dp[i][j] == -1) continue;
            for(int k = 0; k < 8; ++k) {
                if(j>>k&1) continue; 
                int tt = cur[k]+len-1;
                if(tt < in[k].size()) gmax(dp[in[k][tt]+1][j | (1<<k)], dp[i][j]);
                if(tt+1 < in[k].size()) gmax(dp[in[k][tt+1]+1][j | (1<<k)], dp[i][j]+1);
            }
        }
       cur[a[i]-1] ++;
    }
    int ans = -1;
    for(int i = 0; i <= n; ++i) gmax(ans, dp[i][255]);
    if(ans == -1) return 0;
    else return ans*(len+1) + (8-ans)*len;
}
int main(){
    while(~scanf("%d",&n)) {
        for(int i = 0; i < 10; ++i) in[i].clear();
        for(int i = 0; i < n; ++i) {
            scanf("%d",&a[i]);
            in[a[i]-1].push_back(i);    
        }
        int l = 1, r = n/8;
        while(l <= r) {
            int mid = (l+r)>>1;
            if(ok(mid)) l = mid+1;
            else r = mid-1;
        }
        int ans = max(ok(l), ok(r));
        if(ans == 0) {
            for(int i = 0; i < 8; ++i) {
                if(!in[i].empty())
                    ans ++;
            }
        }
        printf("%d\n",ans);
    }
    return 0;
}