• 【CF】323 Div2. D. Once Again...


    挺有意思的一道题目。
    考虑长度为n的数组,重复n次,可以得到n*n的最长上升子序列。同理,也可以得到n*n的最长下降子序列。
    因此,把t分成prefix(上升子序列) + cycle(one integer repeating) + sufix(下降子序列)。
    当t<=2*n时,暴力解。
    注意数据范围。

      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
    ", 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 prefix
    100     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 suffix
    112     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     // iteration
    122     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
    ", ans);
    134     
    135     #ifndef ONLINE_JUDGE
    136         printf("time = %d.
    ", (int)clock());
    137     #endif
    138     
    139     return 0;
    140 }
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  • 原文地址:https://www.cnblogs.com/bombe1013/p/4863850.html
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