[hdu4513]常规dp

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)。

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/* ******************************************************************************** */ #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;                                                                       // }                                                                                   //                                                                                     //                                                                                     //                                                                                     // /* ******************************************************************************** */