【CF】323 Div2. D. Once Again...

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挺有意思的一道題目。
考慮長度為n的數組，重複n次，可以得到n*n的最長上升子序列。同理，也可以得到n*n的最長下降子序列。
因此，把t分成prefix（上升子序列) + cycle(one integer repeating) + sufix(下降子序列)。
當t<=2*n時，暴力解。
注意資料範圍。

  1 /* 583D */  2 #include <iostream>  3 #include <string>  4 #include <map>  5 #include <queue>  6 #include <set>  7 #include <stack>  8 #include <vector>  9 #include <deque> 10 #include <algorithm> 11 #include <cstdio> 12 #include <cmath> 13 #include <ctime> 14 #include <cstring> 15 #include <climits> 16 #include <cctype> 17 #include <cassert> 18 #include <functional> 19 #include <iterator> 20 #include <iomanip> 21 using namespace std; 22 //#pragma comment(linker,"/STACK:102400000,1024000") 23  24 #define sti                set<int> 25 #define stpii            set<pair<int, int> > 26 #define mpii            map<int,int> 27 #define vi                vector<int> 28 #define pii                pair<int,int> 29 #define vpii            vector<pair<int,int> > 30 #define rep(i, a, n)     for (int i=a;i<n;++i) 31 #define per(i, a, n)     for (int i=n-1;i>=a;--i) 32 #define clr                clear 33 #define pb                 push_back 34 #define mp                 make_pair 35 #define fir                first 36 #define sec                second 37 #define all(x)             (x).begin(),(x).end() 38 #define SZ(x)             ((int)(x).size()) 39 #define lson            l, mid, rt<<1 40 #define rson            mid+1, r, rt<<1|1 41  42 const int maxn = 105; 43 const int maxm = 305; 44 int c[maxm]; 45 int a[maxn], b[maxn*maxn*2]; 46 int dp[maxm]; 47 int suf[maxn*maxn], pre[maxn*maxn]; 48  49 int main() { 50     ios::sync_with_stdio(false); 51     #ifndef ONLINE_JUDGE 52         freopen("data.in", "r", stdin); 53         freopen("data.out", "w", stdout); 54     #endif 55      56     int n, t; 57     int mx; 58     int ans = 0; 59      60     scanf("%d %d", &n, &t); 61     rep(i, 1, n+1) { 62         scanf("%d", &a[i]); 63         ++c[a[i]]; 64     } 65      66     if (t <= n*2) { 67         rep(i, 1, n+1) { 68             int k = i; 69             rep(j, 1, t+1) { 70                 b[k] = a[i]; 71                 k += n; 72             } 73         } 74          75         int n_ = n*t; 76          77         memset(dp, 0, sizeof(dp)); 78         rep(i, 1, n_+1) { 79             mx = 0; 80             rep(j, 1, b[i]+1) 81                 mx = max(mx, dp[j]); 82             pre[i] = ++mx; 83             dp[b[i]] = mx; 84         } 85         rep(i, 1, n_+1) 86             ans = max(ans, pre[i]); 87         printf("%d\n", ans); 88         return 0; 89     } 90      91     rep(i, 1, n+1) { 92         int k = i; 93         rep(j, 1, n+1) { 94             b[k] = a[i]; 95             k += n; 96         } 97     } 98      99     // calculate prefix100     int n_ = n*n;101     102     memset(dp, 0, sizeof(dp));103     rep(i, 1, n_+1) {104         mx = 0;105         rep(j, 1, b[i]+1)106             mx = max(mx, dp[j]);107         pre[i] = ++mx;108         dp[b[i]] = mx;109     }110     111     // calculate suffix112     memset(dp, 0, sizeof(dp));113     per(i, 1, n_+1) {114         mx = 0;115         rep(j, b[i], maxm)116             mx = max(mx, dp[j]);117         suf[i] = ++mx;118         dp[b[i]] = mx;119     }120     121     // iteration122     int tmp, n2 = n+n;123     124     rep(i, 1, n+1) {125         rep(j, 1, n+1) {126             if (a[j] < a[i])127                 continue;128             tmp = pre[i+n_-n] + suf[j] + c[a[i]]*(t-n2);129             ans = max(ans, tmp);130         }131     }132     133     printf("%d\n", ans);134     135     #ifndef ONLINE_JUDGE136         printf("time = %d.\n", (int)clock());137     #endif138     139     return 0;140 }

 

【CF】323 Div2. D. Once Again...