CF967D Resource Distribution

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：

 #include <bits/stdc++.h>
 using namespace std;
 typedef pair<int, int> pii;
 pii c[];
 int n, x1, x2;
 vector<int> check(int a, int b)
 {
     bool flg = false;
     int tmp, tmp2, l, l2, i = ;
     for ( ; i < n; i++)
     {
         tmp = c[i].first; l = (a + tmp - ) / tmp;
         if (i + l >= n) continue;
         tmp2 = c[i + l].first; l2 = (b + tmp2 - ) / tmp2;
         if (i + l + l2 > n) continue;
         flg = true; break;
     }
     vector<int> ans;
     if (flg) { ans.push_back(i); ans.push_back(l); ans.push_back(l2); }
     return ans;
 }
 void show(vector<int> & v, bool rev)
 {
     vector<int> v1, v2;
     for (int i = v[]; i < v[] + v[]; i++) v1.push_back(c[i].second);
     for (int i = v[] + v[]; i < v[] + v[] + v[]; i++) v2.push_back(c[i].second);
     cout << "Yes" << endl;
     if (rev)
     {
         cout << v[] << " " << v[] << endl;
         for (auto it: v2) cout << it << " ";
         cout << endl;
         for (auto it: v1) cout << it << " ";
         cout << endl;
     }
     else
     {
         cout << v[] << " " << v[] << endl;
         for (auto it: v1) cout << it << " ";
         cout << endl;
         for (auto it: v2) cout << it << " ";
         cout << endl;
     }
 }
 int main()
 {
     while (cin >> n >> x1 >> x2)
     {
         for (int i = ; i < n; i++) { cin >> c[i].first; c[i].second = i + ; }
         sort(c, c + n);
         vector<int> ans = check(x1, x2);
         if (ans.size()) { show(ans, false); continue; }
         ans = check(x2, x1);
         if (ans.size()) { show(ans, true); continue; }
         cout << "No" << endl;
     }
     return ;
 }