2019牛客暑期多校训练营（第六场） Move

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

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 v_i. 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 \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 \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 v_1 , v_2, …, v_n

(1 \leq v_1, v_2, \ldots, v_n \leq 10001≤v_i <= 1000), where the i-th integer v_i 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.

输入

1
5 3
1 2 3 4 5

输出

Case #1: 5

Brief description

将n个物品放在k个等体积的箱子里，每次放的策略是优先选最大能放进箱子里的物品
问箱子的最少体积是多少

Solution

上来就二分，先WA了一发后，想到解可能没有单调性。并不是体积越大越好。
那没办法，只能改为挨个遍历
发现结果一定在区间【(ceil(sum / k)), (ceil(sum / k) + maxVol)】中
sum为物品重量和，maxVol为最重物品的重量

WA Code

#include <bits/stdc++.h>#define si(a) scanf("%d", &a)
#define sii(a, b) scanf("%d", &a, &b)
#define siii(a, b, c) scanf("%d", &a, &c)
#define siiii(a, b, c, d) scanf("%d", &a, &c, &d)
#define pi(a) printf("%d\n", a)
#define pii(a, b) printf("%d %d\n", a, b)
#define piii(a, b, c) printf("%d %d% d\n", a, b, c)
#define piiii(a, b, c, d) printf("%d %d %d %d\n", a, b, c, d)
#define forRange(i, j, k) for(int i = j; i < k; ++i)
#define mem(a, b) memset(a, b, sizeof(a))
#define mCopy(to, from) memcpy(to, from, sizeof(from))
#define testData(f) freopen(f,"r",stdin)
using namespace std;
const int maxn = 1005;
const int INF = 0x3f3f3f3f;
int n, k;
int volume[maxn];
bool used[maxn];bool test(int boxVol) {mem(used, 0);int currBox = 1, currVolLeft = boxVol;while (currBox <= k) {for (int i = n - 1; i >= 0; --i) {if (!used[i]) {if (volume[i] <= currVolLeft) {currVolLeft -= volume[i];used[i] = true;}}}++currBox;currVolLeft = boxVol;}for (int i = 0; i < n; ++i)if (!used[i])return false;return true;
}int main() {ios::sync_with_stdio(false);cin.tie(NULL);int t;cin >> t;for (int i = 1; i <= t; ++i) {cin >> n >> k;int sum = 0, maxVol = -1;for (int j = 0; j < n; ++j) {cin >> volume[j];sum += volume[j];maxVol = max(maxVol, volume[j]);}sort(volume, volume + n);int l = maxVol, r = sum;while (l <= r) {int mid = (l + r) >> 1;if (test(mid))r = mid - 1;else l = mid + 1;}cout << "Case #" << i << ": " << l << "\n";}fflush(stdout);
}

AC Code

#include <bits/stdc++.h>#define si(a) scanf("%d", &a)
#define sii(a, b) scanf("%d", &a, &b)
#define siii(a, b, c) scanf("%d", &a, &c)
#define siiii(a, b, c, d) scanf("%d", &a, &c, &d)
#define pi(a) printf("%d\n", a)
#define pii(a, b) printf("%d %d\n", a, b)
#define piii(a, b, c) printf("%d %d% d\n", a, b, c)
#define piiii(a, b, c, d) printf("%d %d %d %d\n", a, b, c, d)
#define forRange(i, j, k) for(int i = j; i < k; ++i)
#define mem(a, b) memset(a, b, sizeof(a))
#define mCopy(to, from) memcpy(to, from, sizeof(from))
#define testData(f) freopen(f,"r",stdin)
using namespace std;
const int maxn = 1005;
const int INF = 0x3f3f3f3f;
int n, k;
int volume[maxn];
bool used[maxn];bool test(int boxVol) {mem(used, 0);int currBox = 1, currVolLeft = boxVol;while (currBox <= k) {for (int i = n - 1; i >= 0; --i) {if (!used[i]) {if (volume[i] <= currVolLeft) {currVolLeft -= volume[i];used[i] = true;}}}++currBox;currVolLeft = boxVol;}for (int i = 0; i < n; ++i)if (!used[i])return false;return true;
}int main() {ios::sync_with_stdio(false);cin.tie(NULL);int t;cin >> t;for (int i = 1; i <= t; ++i) {cin >> n >> k;int sum = 0, maxVol = -1;for (int j = 0; j < n; ++j) {cin >> volume[j];sum += volume[j];maxVol = max(maxVol, volume[j]);}sort(volume, volume + n);int l = static_cast<int>(ceil(sum / k)), r = static_cast<int>(ceil(sum / k) + maxVol);int ans = 1;for (int j = l; j <= r; ++j)if (test(j)) {ans = j;break;}cout << "Case #" << i << ": " << ans << "\n";}fflush(stdout);
}