A set of n 1-dimensional items have to be packed in identical bins. All bins have exactly the same length l and each item i has length li ≤ l. We look for a minimal number of bins q such that • each bin contains at most 2 items, • each item is packed in one of the q bins, • the sum of the lengths of the items packed in a bin does not exceed l. You are requested, given the integer values n, l, l1, . . . , ln, to compute the optimal number of bins q. Input The input begins with a single positive integer on a line by itself indicating the number of the cases following, each of them as described below. This line is followed by a blank line, and there is also a blank line between two consecutive inputs. The first line of the input file contains the number of items n (1 ≤ n ≤ 105 ). The second line contains one integer that corresponds to the bin length l ≤ 10000. We then have n lines containing one integer value that represents the length of the items. Output For each test case, your program has to write the minimal number of bins required to pack all items. The outputs of two consecutive cases will be separated by a blank line. Note: The sample instance and an optimal solution is shown in the figure below. Items are numbered from 1 to 10 according to the input order. Sample Input 1 10 80 70 15 30 35 10 80 20 35 10 30 Sample Output 6
思路：先对数据排序，对每一个目前最大值找到可以最大程度匹配的值，若找不到，则自己装到一个背包
#include <iostream> #include<algorithm> using namespace std; int a[100005]; int main() { int t, n, max, sum; cin >> t; while (t--) { int a[100001]; sum = 0; cin >> n >> max; for (int i = 0; i < n; i++) cin >> a[i]; sort(a, a + n); for (int i = n - 1; i >= 0; i--) { if (a[i] == 0) continue; for (int j = i - 1; j >= 0; j--) { if (a[j] == 0) continue; if (a[i] + a[j] <= max) { a[j] = 0; break; } } a[i] = 0; sum++; } cout << sum << endl; if (t) cout << endl; } return 0; }