Ice Cream Tower（The 2016 ACM-ICPC Asia China-Final Contest 二分&贪心）

题目：

　　Mr. Panda likes ice cream very much especially the ice cream tower. An ice cream tower consists of K ice cream balls stacking up as a tower. In order to make the tower stable, the lower ice cream ball should be at least twice as large as the ball right above it. In other words, if the sizes of the ice cream balls from top to bottom are A₀, A₁, A₂, · · · , A_K−1, then A₀ × 2 ≤ A₁, A₁ × 2 ≤ A₂, etc.

　　One day Mr. Panda was walking along the street and found a shop selling ice cream balls. There are N ice cream balls on sell and the sizes are B₀, B₁, B₂, · · · , B_N−1. Mr. Panda was wondering the maximal number of ice cream towers could be made by these balls.

Input：

　　The first line of the input gives the number of test cases, T. T test cases follow. Each test case starts with a line consisting of 2 integers, N the number of ice cream balls in shop and K the number of balls needed to form an ice cream tower. The next line consists of N integers representing the size of ice cream balls in shop.

Output：

　　For each test case, output one line containing “Case #x: y”, where x is the test case number (starting from 1) and y is the maximal number of ice cream towers could be made.

题意：现在给出N个冰激凌球的尺寸和k个球才能摞起一个冰激凌，为冰激凌的稳定性，要求下边的球的尺寸必须至少是上边球尺寸的两倍大。问最多能摞几个冰激凌。

思路：先对给出的尺寸从小到大排一下序，然后从0到n/k进行二分最多能摞x个，二分的答案用贪心来进行检验x能不能得出。

贪心：要想得到当前答案下的最优解，那么排序后的最前边的x个一定是作为冰激凌的顶层球的，而摞成x个冰激凌需要k*x个球，所以循环x*k次看看能不能每次都能找出来，如果每次都能找出来，x就成立，反之不成立。

ps：二分搜索的写法好多门道啊，还得深入学习啊！！

代码：

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <queue>
#include <map>
#include <set>
#include <vector>
using namespace std;
typedef long long ll;
const int maxn = 3e5+10;
ll a[maxn],b[maxn];
int n,k;

bool judge(int x)
{
    for(int i = 0; i<x; i++)
        b[i] = a[i];
    int p = x;
    for(int i = x; i<x*k; i++)//总共需要x*k个球，循环这些次，看每次是不是都能找到
    {
        while(a[p]<b[i-x]*2 && p<n)p++;
        if(p==n)return false;//没有凑齐x*k个
        b[i] = a[p];
        p++;
    }
    return true;
}


int main()
{
    int T,cnt=1;
    scanf("%d",&T);
    while(T--)
    {
        memset(a,0,sizeof(a));
        scanf("%d%d",&n,&k);
        for(int i = 0; i<n; i++)
            scanf("%lld",&a[i]);
        int l = 0, r = n/k;
        sort(a,a+n);
        while(l < r)
        {
            int mid = (l+r+1)/2;
            if(judge(mid))
                l = mid;
            else
                r = mid-1;
        }
        printf("Case #%d: %d\n",cnt++,l);
    }
    return 0;
}
/*
样例输入：
3
4 2
1 2 3 4
6 3
1 1 2 2 4 4
6 3
1 1 2 2 3 4
样例输出：
Case #1: 2
Case #2: 2
Case #3: 1
*/

View Code