Codeforces732E Sockets

首先检测有木有和Computer匹配的Socket，如果有则将其匹配。

然后将所有没有匹配的Socket连上Adapter，再去检测有木有Computer与Socket匹配。

重复这个操作31次，所有Socket的power都变成1了，如果再不能匹配就结束程序。

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3
 4 struct PC
 5 {
 6     int power;
 7     int idx;
 8     friend bool operator< (const PC& a, const PC& b)
 9     {
10         return a.power < b.power;
11     }
12 };
13 int s[200010];
14 bool vis[200010];
15 int a[200010];
16 int b[200010];
17
18 int main()
19 {
20     int n, m;
21     scanf("%d%d", &n, &m);
22     multiset<PC> pc;
23     int p;
24     for (int i = 1; i <= n; i++)
25     {
26         scanf("%d", &p);
27         pc.insert({p, i});
28     }
29     for (int i = 1; i <= m; i++)
30         scanf("%d", s + i);
31     int c = 0, u = 0;
32     for (int i = 0; i < 31; i++)
33     {
34         for (int j = 1; j <= m; j++)
35         {
36             if (!vis[j])
37             {
38                 multiset<PC>::iterator it = pc.find({s[j], 0});
39                 if (it != pc.end())
40                 {
41                     b[it->idx] = j;
42                     vis[j] = true;
43                     pc.erase(it);
44                     c++;
45                     u += a[j];
46                 }
47             }
48         }
49         for (int j = 1; j <= m; j++)
50             if (!vis[j])
51                 a[j]++, s[j] = (s[j] + 1) / 2;
52     }
53     printf("%d %d\n", c, u);
54     for (int i = 1; i <= m; i++)
55         printf("%d ", vis[i] ? a[i] : 0);
56     puts("");
57     for (int i = 1; i <= n; i++)
58         printf("%d ", b[i]);
59     return 0;
60 }