Codeforces Round #385 (Div. 2) E

二分答案+状压DP。

状态转移写崩了，查了好久，对于每一个点是可选可不选的，我漏了一个不选的状态，wa了好久。

思路：重复的个数肯定是n和n-1，因此二分n，对于每一个n进行一次状压DP。在此之前先打一个f数组用于以后的状态转移。

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
#include <stack>
#include <string>
using namespace std;
const int maxn = 1005;
int f[maxn][9][maxn];
int dp[maxn][1 << 8];
int a[maxn];
int ans = 0;
string tobinary(int x)
{
stack<int> s;
string ret;
do
{
s.push(x % 2);
x /= 2;
}while(x);
while(!s.empty())
{
ret.push_back('0' + s.top());
s.pop();
}
return ret;
}
int main()
{
memset(f, -1, sizeof(f));
int n;
scanf("%d", &n);
for(int i = 0; i < n; ++i)
scanf("%d", &a[i]);
for(int i = 0; i < n; ++i)
{
for(int j = 1; j <= 8; ++j)
{
int sum = 0;
for(int k = i; k < n; ++k)
{
if(a[k] == j)
{
f[i][j][++sum] = k;
}
}
}
}
int l = 1, r = n;
while(l < r)
{
int mid = (l + r + 1) >> 1;
memset(dp, 0, sizeof(dp));
for(int i = 0; i < n; ++i)
{
int x = a[i];
for(int j = 0; j < (1 << 8); ++j)
{
if((((1 << (x - 1)) & j) == 0) && dp[i][j] != 0 || j == 0)
{
int id1 = f[i][x][mid];
int id2 = f[i][x][mid - 1];
if(id1 != -1)
dp[id1 + 1][(1 << (x - 1)) | j] = max(dp[id1 + 1][(1 << (x - 1)) | j], dp[i][j] + mid);
if(id2 != -1)
dp[id2 + 1][(1 << (x - 1)) | j] = max(dp[id2 + 1][(1 << (x - 1)) | j], dp[i][j] + mid - 1);
// if(mid == 2)
// {
// cout << tobinary((1 << (x - 1)) | j) << " ";
// cout << "id1 == " << id1 << " id2 == " << id2 << " ";
// cout << "dp[" << i << "][" << tobinary(j) << "] == " << dp[i][j] << endl;
// }
}
dp[i + 1][j] = max(dp[i + 1][j], dp[i][j]);
}
}
int flag = 0;
for(int i = 1; i <= n; ++i)
{
if(dp[i][(1 << 8) - 1] > 0)
{
ans = max(ans, dp[i][(1 << 8) - 1]);
flag = 1;
// cout << "i == " << i << endl;
// cout << dp[i][(1 << 8) - 1] << endl;
}
}
if(flag)
{
l = mid;
}
else
{
r = mid - 1;
}
// cout << "mid == " << mid << " flag == " << flag << endl;
}
if(l == 1)
{
int vis[9];
memset(vis, 0, sizeof(vis));
for(int i = 0; i < n; ++i)
{
int x = a[i];
if(!vis[x])
{
++ans;
vis[x] = 1;
}
}
}
printf("%d\n", ans);
}