2019牛客暑期多校训练营（第六场）D Move(表面二分实则暴力)

链接：https://ac.nowcoder.com/acm/contest/886/D
来源：牛客网
 

时间限制：C/C++ 2秒，其他语言4秒
空间限制：C/C++ 262144K，其他语言524288K
64bit IO Format: %lld

题目描述

After the struggle of graduating from college, TangTang is about to move from a student apartment to his new home.

TangTang has n items to move, the i-th of which is of volume viv_ivi​. He can pack all these items into at most K boxes of the same volume.
 

TangTang is so clever that he uses the following strategies for packing items:

- Each time, he would put items into a box by the next strategy, and then he would try to fill another box.

- For each box, he would put an unpacked item of the largest suitable volume into the box repeatedly until there is no such item that can be fitted in the box.

Now, the question is what is the minimum volume of these boxes required to pack all items.

输入描述:

There are multiple test cases. The first line contains an integer T (1≤T≤201 \leq T \leq 201≤T≤20), indicating the number of test cases. Test cases are given in the following.For each test case, the first line contains two integers n, K (1≤n,K≤10001 \leq n, K \leq 10001≤n,K≤1000), representing the number of items and the number of boxes respectively.The second line contains n integers v1v_1v1​, v2v_2v2​, …\ldots…, vnv_nvn​ (1≤v1,v2,…,vn≤10001 \leq v_1, v_2, \ldots, v_n \leq 10001≤v1​,v2​,…,vn​≤1000), where the i-th integer viv_ivi​ represents the volume of the i-th item.

输出描述:

For each test case, output "Case #x: y" in one line (without quotes), where x indicates the case number starting from 1, and y denotes the answer to this test case.

示例1

输入

复制

1
5 3
1 2 3 4 5

输出

复制

Case #1: 5

                没想到没想到。。。这题非常巧妙地一点就是让人一眼觉得这是二分直接做，但是是事实上我们二分的（包裹的容量）不是单调的量~在某一个成立的时候下一个点可能会失败。。。比如：

              这种情况下我们从另一个方面考虑：我们之前二分的量（包裹的容量）的下界肯定是（sum/k）(sum为物体容量和)，而他的上界即是(sum/k+Wmax)(Wmax为物体容量的最大值)保证要放肯定能放上~这样的话我们可以得出即使枚举（包裹的容量）也只是n的大小，枚举的次数根本不多。。然后就这么暴力做就行了~

#include<bits/stdc++.h>
using namespace std;
int da[1005]; int n, m, all;
bool check(int mid) {if (m*mid < all)return 0;vector<int>q;for (int i = 1; i <= n; i++) {q.push_back(da[i]);}int sz = n; int sum = 0, su = 1;q.push_back(0x3f3f3f3f);sort(q.begin(), q.end());while (q.size() != 1) {int need = mid - sum;int pos = upper_bound(q.begin(), q.end(), need) - q.begin();if (pos <= 0) {sum = 0; su++;continue;}pos--;sum += q[pos];q.erase((q.begin() + pos));}if (sum == 0)su--;return su <= m;
}
int main() {ios::sync_with_stdio(0);vector<int>q; q.push_back(1); q.push_back(0);int te;cin >> te; int cas = 1;while (te--) {cin >> n >> m;all = 0;int l = -0x3f3f3f3f;for (int i = 1; i <= n; i++) {cin >> da[i]; all += da[i];l = max(l, da[i]);}int r = l + all / m;l = all / m;int ans;for (int i = l; i <= r; i++) {if (check(i)) {ans = i;break;}}cout << "Case #" << cas << ": " << ans << "\n"; cas++;}return 0;
}