hdu 2192 MagicBuilding

MagicBuilding

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1224    Accepted Submission(s): 551

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

212 51 2 2 3 3

 

Sample Output

12

 

#include<stdio.h>#include <iostream>using namespace std;struct node{    int a;    bool flag;   //标记是否合并};int cmp(const void *a,const void *b){    node *d1=(node*)a;    node *d2=(node*)b;    return (d2->a)-(d1->a);}int main(){    node d[10001];    int n,k,N;    int ans,prew;    cin>>N;    while (N--)    {        k=0;        ans=0;        cin>>n;        for (int i=0; i<n; i++)        {            cin>>d[i].a;            d[i].flag=true;        }        qsort(d, n, sizeof(node), cmp);         while (k<n)        {            prew=-1;            for (int i=0; i<n; i++)            {                if(d[i].flag && d[i].a!=prew)                {                    d[i].flag=false;                    prew=d[i].a;                    k++;                }            }            ans++;        }        cout<<ans<<endl;    }    return 0;}