• BZOJ3308 九月的咖啡店


    Orz PoPoQQQ

    话说这题还有要注意的地方。。。

    就是。。。不能加SLF优化,千万不能加

    n = 40000,不加本机跑出来2sec,加了跑出来40sec。。。【给跪了

      1 /**************************************************************
      2     Problem: 3308
      3     User: rausen
      4     Language: C++
      5     Result: Accepted
      6     Time:27500 ms
      7     Memory:32752 kb
      8 ****************************************************************/
      9  
     10 #include <cstdio>
     11 #include <algorithm>
     12 #include <bitset>
     13  
     14 using namespace std;
     15 const int N = 2e4 + 5;
     16 const int Cnt = N * 10;
     17 const int M = Cnt * 10;
     18 const int inf = 1e9;
     19  
     20 struct edges {
     21     int next, to, f, cost;
     22     edges() {}
     23     edges(int _n, int _t, int _f, int _c) : next(_n), to(_t), f(_f), cost(_c) {}
     24 } e[M];
     25  
     26 int n, ans, S, T;
     27 int first[N], tot = 1;
     28 bitset <Cnt> not_p;
     29 int pr[N], tot_p, tot_a, tot_b;
     30 int d[N], g[N], v[N];
     31  
     32 inline void Add_Edges(int x, int y, int f, int c) {
     33     e[++tot] = edges(first[x], y, f, c), first[x] = tot;
     34     e[++tot] = edges(first[y], x, 0, -c), first[y] = tot;
     35 }
     36  
     37 inline void calc() {
     38     static int x;
     39     for (x = g[T]; x; x = g[e[x ^ 1].to])
     40         --e[x].f, ++e[x ^ 1].f;
     41 }
     42  
     43 #define y e[x].to
     44 inline bool spfa() {
     45     static int x, now, q[70000];
     46     static unsigned short l, r;
     47     for (x = 1; x <= T; ++x)
     48         d[x] = -inf;
     49     d[S] = 0, v[S] = 1, q[0] = S;
     50     for(l = r = 0; l != r + 1; ) {
     51         now = q[l++];
     52         for (x = first[now]; x; x = e[x].next) {
     53             if (e[x].f && d[now] + e[x].cost > d[y]) {
     54                 d[y] = d[now] + e[x].cost, g[y] = x;
     55                 if (!v[y])
     56                     v[y] = 1, q[++r] = y;
     57             }
     58         }
     59         v[now] = 0;
     60     }
     61     return d[T] >= 0;
     62 }
     63 #undef y
     64  
     65 inline int work() {
     66     int res = 0;
     67     while (spfa())
     68         calc(), res += d[T];
     69     return res;
     70 }
     71  
     72 void get_prime(int N) {
     73     int i, j, k;
     74     for (i = 2; i <= n; ++i) {
     75         if (!not_p[i]) pr[++tot_p] = i;
     76         for (j = 1; j <= tot_p; ++j) {
     77             if ((k = i * pr[j]) > N) break;
     78             not_p[k] = 1;
     79             if (i % pr[j] == 0) break;
     80         }
     81     }
     82 }
     83  
     84 inline int get(int n, int p) {
     85     static int res;
     86     for (res = 1; res * p <= n; res *= p);
     87     return res;
     88 }
     89  
     90 int main() {
     91     int i, j, tmp;
     92     scanf("%d", &n);
     93     get_prime(n);
     94     for (i = 1; i <= tot_p && 1ll * pr[i] * pr[i] <= n; ++i) ++tot_a;
     95     for (; i <= tot_p && pr[i] * 2 <= n; ++i) ++tot_b;
     96     for (; i <= tot_p; ++i) ans += pr[i];
     97     S = tot_a + tot_b + 1, T = S + 1;
     98  
     99 #define J j + tot_a
    100     for (i = 1; i <= tot_a; ++i)
    101         Add_Edges(S, i, 1, 0), Add_Edges(i, T, 1, get(n, pr[i]));
    102     for (j = 1; j <= tot_b; ++j)
    103         Add_Edges(S, J, 1, pr[J]), Add_Edges(J, T, 1, 0);
    104     for (i = 1; i <= tot_a; ++i) 
    105         for (j = 1; j <= tot_b; ++j)
    106             if ((tmp = get(n / pr[J], pr[i]) * pr[J]) > get(n, pr[i]) + pr[J])
    107                 Add_Edges(i, J, 1, tmp);
    108 #undef J
    109     printf("%d
    ", ans + 1 + work());
    110     return 0;
    111 }
    View Code

    (p.s. 成功成为最慢的2333)

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  • 原文地址:https://www.cnblogs.com/rausen/p/4507582.html
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