hdu 5162 Jump and Jump...（模拟）

Jump and Jump...

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 1500    Accepted Submission(s): 787


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

题意：n个孩子，每个孩子跳3下，最远的一下为孩子的跳远距离，输入保证不会有两个孩子的最远跳远距离相等。

输出1~n号孩子的排名

思路：直接模拟即可

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
struct Node
{
int id,dis;
}kid[10];
bool cmp(Node a,Node b)
{
return a.dis>b.dis;
}
int ans[10];
int main()
{
int T,n;
int dis,dis1,dis2;
scanf("%d",&T);
while(T--)
{
scanf("%d",&n);
for(int i=1;i<=n;i++)
{
scanf("%d %d %d",&dis,&dis1,&dis2);
dis=max(dis,dis1);
kid[i].dis=max(dis,dis2);
kid[i].id=i;
}
sort(kid+1,kid+1+n,cmp);
for(int i=1;i<=n;i++)
ans[kid[i].id]=i;
for(int i=1;i<n;i++)
printf("%d ",ans[i]);
printf("%d\n",ans
);
}
return 0;
}