Uva1149 Bin Packing

Uva1149 Bin Packing

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

Solution

这是一道基础的贪心水题，我们可以将物品按重量从小到大排序，每次将当前最轻的物品与能与其放在一起的最重的物品放在一起。

• 代码

#include <algorithm> #include <cstdio>#include <cstdlib>using namespace std;const int Inf = 2147483647;const int maxn = 100000 + 5;int n, m, ans, L[maxn];inline int read(){    char c;    do {      c = getchar();    }while(c < '0' || c > '9');    int sum = 0;    do {      sum = sum * 10 + c - 48;      c = getchar();    }while(c >= '0' && c <= '9');    return sum;}int main(){    freopen("input.in", "r", stdin);    freopen("output.out", "w", stdout);    int T, i, j;    T = read();    while(T--) {      n = read(); m = read();      for(i = 1; i <= n; i++)        L[i] = read();      sort(&L[1], &L[n + 1]);      ans = 0; //记得清0      for(i = 1, j = n; i <= j;) {        if(L[i] + L[j] > m) {          j--; ans++;        }else {          i++; j--; ans++;          }      }      printf("%d\n", ans);      if(T) printf("\n");    }                                                                  return 0;}