Bin Packing - UVALive 3503 贪心

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Bin Packing

Time Limit:3000MS Memory Limit:0KB 64bit IO Format:%lld & %llu

Description

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 l_i $\le$ 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 , ..., l_n , 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 $\le$ n $\le$ 10₅) . The second line contains one integer that corresponds to the bin length l $\le$ 10000 . We then have n lines containing one integer value that represents the length of the items.

Output

For each test case, the output must follow the description below. The outputs of two consecutive cases will be separated by a blank line.

For each input file, your program has to write the minimal number of bins required to pack all items.

Sample Input

1108070153035108020351030

Sample Output

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.

$\epsfbox{p3503.eps}$

题意：让最多两块组合起来，并且限定最大长度，使得组数最少。

思路：让最长的和最短的匹配。

AC代码如下：

#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>#include<cmath>using namespace std;int num[100010];int main(){ int T,t,n,len,i,j,k,ans,l,r;  scanf("%d",&T);  for(t=1;t<=T;t++)  { scanf("%d%d",&n,&len);    ans=0;    for(i=1;i<=n;i++)     scanf("%d",&num[i]);    sort(num+1,num+1+n);    l=1;r=n;    while(l<r)    { if(num[l]+num[r]<=len)      { ans++;        l++;        r--;      }      else      { ans++;        r--;      }    }    if(l==r)     ans++;    if(t!=1)     printf("\n");    printf("%d\n",ans);  }}

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