hdu---2192MagicBuilding

最新推荐文章于 2019-04-02 21:44:00 发布

weixin_30248399

最新推荐文章于 2019-04-02 21:44:00 发布

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文章标签： java 人工智能

原文链接：http://www.cnblogs.com/songacm/p/3538447.html

本文针对MagicBuilding问题，提出了一种有效的解决策略。通过对建筑物大小进行快速排序，算法能够确定最少数量的MagicBuilding组合，以此来节约空间。适用于人口密集地区，如MagicStar，通过将不同大小的建筑群集起来实现土地资源的有效利用。

MagicBuilding

Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1067 Accepted Submission(s): 478

Problem Description

As the increase of population, the living space for people is becoming smaller and smaller. In MagicStar the problem is much worse. Dr. Mathematica is trying to save land by clustering buildings and then we call the set of buildings MagicBuilding. Now we can treat the buildings as a square of size d, and the height doesn't matter. Buildings of d1,d2,d3....dn can be clustered into one MagicBuilding if they satisfy di != dj(i != j).
Given a series of buildings size , you need to calculate the minimal numbers of MagicBuildings that can be made. Note that one building can also be considered as a MagicBuilding.
Suppose there are five buildings : 1, 2, 2, 3, 3. We make three MagicBuildings (1,3), (2,3), (2) .And we can also make two MagicBuilding :(1,2,3), (2,3). There is at least two MagicBuildings obviously.

Input

The first line of the input is a single number t, indicating the number of test cases.
Each test case starts by n (1≤n≤10^4) in a line indicating the number of buildings. Next n positive numbers (less than 2^31) will be the size of the buildings.

Output

For each test case , output a number perline, meaning the minimal number of the MagicBuilding that can be made.

Sample Input

2 1 2 5 1 2 2 3 3

Sample Output

1 2

//题目描述就不说了，因为n最大为10000，所以直接快排，再查找即可。 (n为10000时，一般做到n^2都不会超时，这里快排只有n*log2n，是肯定可行的。)

代码：

#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
#include <algorithm>
using namespace std;
#define N 10010
int f[N];
int cmp(int x,int y){
    if(x>y) return 1;
    else return 0;
}
int main()
{
    freopen("in.txt","r",stdin);
    int i,j,a,b,m,n;int len,T;
    scanf("%d",&T);
    while(T--){
        scanf("%d",&n);
        for(i=0;i<n;i++)
            scanf("%d",&f[i]);
        sort(f,f+n);
        int num=1;int max=1;
        for(i=1;i<n;i++){
            if(f[i]==f[i-1]) num++;
            else{
                if(num>max) max=num;
                num=1;
            }
        }
        if(num>max) max=num;
        printf("%d\n",max);
    }

    return 0;
}
//其中for循环可以改成下面的形式，最终直接输出len即可！
for (i = 1; i < n;++i)
    if (a[i] == a[i-len])
        len++;

转载于:https://www.cnblogs.com/songacm/p/3538447.html