CF967D Resource Distribution

Posted 2020-11-02 王宜鸣

思路：

在一堆服务器中，资源最少的那一个是“瓶颈”，由此想到贪心思路。

首先对所有服务器按照资源数量c排序，再从头到尾扫描。对每个位置，根据x1和x2计算出两段连续的服务器集合分别分配给A任务和B任务（还需要枚举分配A和B的先后顺序），如果够用，则有解。所有位置都不行，则无解。

实现：

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 typedef pair<int, int> pii;
 4 pii c[300005];
 5 int n, x1, x2;
 6 vector<int> check(int a, int b)
 7 {
 8     bool flg = false;
 9     int tmp, tmp2, l, l2, i = 0;
10     for ( ; i < n; i++)
11     {
12         tmp = c[i].first; l = (a + tmp - 1) / tmp;
13         if (i + l >= n) continue;
14         tmp2 = c[i + l].first; l2 = (b + tmp2 - 1) / tmp2;
15         if (i + l + l2 > n) continue;
16         flg = true; break;
17     }
18     vector<int> ans;
19     if (flg) { ans.push_back(i); ans.push_back(l); ans.push_back(l2); }
20     return ans;
21 }
22 void show(vector<int> & v, bool rev)
23 {
24     vector<int> v1, v2;
25     for (int i = v[0]; i < v[0] + v[1]; i++) v1.push_back(c[i].second);
26     for (int i = v[0] + v[1]; i < v[0] + v[1] + v[2]; i++) v2.push_back(c[i].second);
27     cout << "Yes" << endl;
28     if (rev)
29     {
30         cout << v[2] << " " << v[1] << endl;
31         for (auto it: v2) cout << it << " ";
32         cout << endl;
33         for (auto it: v1) cout << it << " ";
34         cout << endl;
35     }
36     else
37     {
38         cout << v[1] << " " << v[2] << endl;
39         for (auto it: v1) cout << it << " ";
40         cout << endl;
41         for (auto it: v2) cout << it << " ";
42         cout << endl;
43     }
44 }
45 int main()
46 {
47     while (cin >> n >> x1 >> x2)
48     {
49         for (int i = 0; i < n; i++) { cin >> c[i].first; c[i].second = i + 1; }
50         sort(c, c + n);
51         vector<int> ans = check(x1, x2);
52         if (ans.size()) { show(ans, false); continue; }
53         ans = check(x2, x1);
54         if (ans.size()) { show(ans, true); continue; }
55         cout << "No" << endl;
56     }
57     return 0;
58 }