Bestcoder 27 # A jump and jump-优快云博客

本文链接：https://blog.youkuaiyun.com/wikioi_bai/article/details/43105493

本文介绍了一种通过三次跳跃成绩确定每位参赛者最远距离，并据此进行排名的算法实现。适用于2到3位参赛者的小型比赛场景。

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Jump and Jump...

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 102 Accepted Submission(s): 72

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

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.

代码：

# include<cstdio>
# include<iostream>
# include<algorithm>

using namespace std;

# define MAX 5

struct node
{
    int id;
    int jump;
    int rk;
}b[MAX];

int a[MAX];

int cmp( const struct node & x,const struct node & y )
{
    return x.jump>y.jump;
}

int cmp2( const struct node & x,const struct node & y )
{
    return x.id < y.id;
}


int main(void)
{
    int t;cin>>t;
    while ( t-- )
    {
        int n;cin>>n;
        int j = 1;
        for ( int t = 1;t <= n;t++ )
        {
            for ( int i = 1;i <= 3;i++ )
                    {
                        cin>>a[i];
                    }
            sort(a+1,a+1+3);
            b[j++].jump = a[3];
            //j++;
        }
    for ( int i = 1;i <= n;i++ )
    {
        b[i].id = i;
    }
        sort(b+1,b+1+n,cmp);
        for ( int i = 1;i <= n;i++ )
        {
            b[i].rk = i;
        }
        sort(b+1,b+1+n,cmp2);
        for ( int i = 1;i <= n-1;i++ )
        {
            cout<<b[i].rk<<" ";
        }
        cout<<b[n].rk<<endl;
        //cout<<endl;



    }


    return 0;
}