SPOJ：Decreasing Number of Visible Box（不错的，背包？贪心？）

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Shadowman loves to collect box but his roommates woogieman and itman don‘t like box and so shadowman wants to hide boxes as many as possible. A box can be kept hidden inside of another box if and only if the box in which it will be held is empty and the size of the box is at least twice as large as the size of the box.

Print the minimum number of box that can be shown.

Input

The input set starts with single line integer T (1<=T<=50) the number of test cases. Then following T cases starts with an integer N (1<=N<=100000) denoting the number of box. The next line contains N space separated positive integer. i-th of them contains a numbers Ai(1<=Ai<=100000) size of the i-th box.

Output

Output the the case number and the minimum number of box that can be shown.

Example

Input:
2
4
1 2 4 8
4
1 3 4 5

Output:
Case 1: 1
Case 2: 3

题意：给定N个盒子，现在每个盒子最多可以装一个盒子到内部，而且要满足外面的盒子体积大于等于内部的盒子的两倍。

思路：开始以为是平衡树，每次在树上找到一个大于等于两倍体积的点，减减。但是复杂度太高，优化看一下也超时。最后又是手欠看了别人的代码。

           方法大概是，先排序，然后新建一个数组q，代表目前最小的盒子的两倍，如果后面的盒子大于这个q，就可以把q装进去，装一次代表答案减一。

            （主要是利用了单调性，贪心地装目前最小的，秒啊。

#include<bits/stdc++.h>
using namespace std;
const int maxn=1000010;
int a[maxn],q[maxn],head,tail;
int main()
{
    int T,N,i,j,ans,Case=0;
    scanf("%d",&T);
    while(T--){
        head=tail=ans=0;
        scanf("%d",&N);
        for(i=1;i<=N;i++) scanf("%d",&a[i]);
        sort(a+1,a+N+1);
        for(i=1;i<=N;i++){
            if(head==tail) ans++;
            else if(q[tail+1]<=a[i]) tail++;
            else ans++;
            q[++head]=a[i]*2;
        }
        printf("Case %d: %d\n",++Case,ans);
    }
    return 0;
}

SPOJ：Decreasing Number of Visible Box（不错的，背包？贪心？）

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原文地址：https://www.cnblogs.com/hua-dong/p/8971727.html