题意：给一个长度为m的序列，从里面选出一些数，相对位置不发生变化，并满足a[i]=a[n-i],a[1]<a[2]<...<a[(n+1)/2],n是数的个数，求最大的n
思路：dp[i][j]表示0~i,j~m的答案，则dp[i][j]=dp[l][r]+1+(i<j)，其中a[i]=a[j]>a[l]=a[r]&&l<i<=j<r。枚举3个变量i,j,r，维护一个l就行了，o(m^3)。
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 | /* ******************************************************************************** */#include <iostream> //#include <cstdio> //#include <cmath> //#include <cstdlib> //#include <cstring> //#include <vector> //#include <ctime> //#include <deque> //#include <queue> //#include <algorithm> //using namespace std; // //#define pb push_back //#define mp make_pair //#define X first //#define Y second //#define all(a) (a).begin(), (a).end() //#define foreach(i, a) for (typeof(a.begin()) it = a.begin(); it != a.end(); it ++) // //void RI(vector<int>&a,int n){a.resize(n);for(int i=0;i<n;i++)scanf("%d",&a[i]);} //void RI(){}void RI(int&X){scanf("%d",&X);}template<typename...R> //void RI(int&f,R&...r){RI(f);RI(r...);}void RI(int*p,int*q){int d=p<q?1:-1; //while(p!=q){scanf("%d",p);p+=d;}}void print(){cout<<endl;}template<typename T> //void print(const T t){cout<<t<<endl;}template<typename F,typename...R> //void print(const F f,const R...r){cout<<f<<", ";print(r...);}template<typename T> //void print(T*p, T*q){int d=p<q?1:-1;while(p!=q){cout<<*p<<", ";p+=d;}cout<<endl;} // //typedef pair<int, int> pii; //typedef long long ll; //typedef unsigned long long ull; // ///* -------------------------------------------------------------------------------- */ //template<typename T>bool umax(T &a, const T &b) { return a >= b? false : (a = b, true);}int a[300], dp[300][300];int main() {#ifndef ONLINE_JUDGE freopen("in.txt", "r", stdin);#endif // ONLINE_JUDGE int T; RI(T); while (T --) { int n; RI(n); RI(a + 1, a + 1 + n); int p[300] = {}; memset(dp, 0, sizeof(dp)); int ans = 0; for (int i = 1; i <= n; i ++) { for (int j = n; j >= i; j --) { if (a[i] != a[j]) continue; dp[i][j] = 1 + (i < j); for (int k = j + 1; k <= n; k ++) { if (a[k] < a[i]) { if (p[a[k]] && p[a[k]] < i) umax(dp[i][j], dp[p[a[k]]][k] + 1 + (i < j)); } } umax(ans, dp[i][j]); } p[a[i]] = i; } cout << ans << endl; } return 0; //} // // // ///* ******************************************************************************** */ |