Jump and Jump... HDU 5162



Jump and Jump...

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 664    Accepted Submission(s): 370

Problem Description

There are n kids and they want to know who can jump the farthest. For each kid, he can jump three times and the distance he jumps is maximum distance amount all the three jump. For example, if the distance of each jump is (10, 30, 20), then the farthest distance he can jump is 30. Given the distance for each jump of the kids, you should find the rank of each kid.

 

Input

There are multiple test cases. The first line of input contains an integer T (1≤T≤100) , indicating the number of test cases. For each test case: The first line contains an integer n (2≤n≤3) , indicating the number of kids. For the next n lines, each line contains three integers ai,bi and ci ( 1≤ai,bi,ci,≤300 ), indicating the distance for each jump of the i -th kid. It's guaranteed that the final rank of each kid won't be the same (ie. the farthest distance each kid can jump won't be the same).

 

Output

For each test case, you should output a single line contain n integers, separated by one space. The i -th integer indicating the rank of i -th kid.

 

Sample Input

   
   
   
   

    
    
    
    2
3
10 10 10
10 20 30
10 10 20
2
3 4 1
1 2 1
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    3 1 2
1 2
   
   
   
   
   
   
   
   


   
   
   
   
   
   
   
   

    
    
    
    
 
     Hint 
    
For the first case, the farthest distance each kid can jump is 10, 30 and 20. So the rank is 3, 1, 2.
   
   
   
   
   
   
   
   

    
    
    
     
   
   
   
   

 

很粗暴的方法，开两个数组，存储好每个人的最远距离，把一个数组排好序后，和另一个个数组比较，得出名次关系

#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;

int T,sum[105],a,b,c,n,i,j,temp[105];

int main()
{
    cin >> T;
    while(T--)
    {
        cin >> n;
        memset(sum,0,sizeof(int)*105);
        memset(temp,0,sizeof(int)*106);
        for(i = 0;i < n;i++)
            {
                cin >> a >>b >> c;
                sum[i] = max(max(a,b),c);
                temp[i] = sum[i];
            }
        for(i = 0;i < n;i++)
            for(j = 0;j < n - 1 - i;j++)
            if(sum[j] < sum[j+1])
        {
            sum[j]+=sum[j + 1];
            sum[j+1] = sum[j] - sum[j+1];
            sum[j] = sum[j] - sum[j+1];
        }
        for(i = 0;i < n;i++)
                for(j = 0;j < n;j++)
                if(sum[j] == temp[i])
                    if(!i) cout << j+1;
                    else cout << " " << j+1;
                cout << endl;
    }
    return 0 ;
}