cf23C Oranges and Apples (贪心_好题)

给你2*n-1个篮子，分别装有若干苹果跟橘子，问取出n个篮子，使得苹果的总量占苹果sum的一半以上，且橘子也占橘子sum二等一半以上

仔细观察数据，你会发现这是一个贪心策略、。。。

我们一开始排序会想到，苹果从大往小排，去前n个，但不能保证橘子能占一半以上...

那么如果有两个橘子，我们当然取最大的最好，那么如果我们每次取橘子最大的，并舍弃另一个，那么橘子能保证了，但是苹果不能保证....

由此，我们想到了苹果从小到大排，那么两者窦娥能保证了....

Description

In 2N - 1 boxes there are apples and oranges. Your task is to choose N boxes so, that they will contain not less than half of
all the apples and not less than half of all the oranges.

Input

The first input line contains one number T — amount of tests. The description of each test starts with a natural number N —
amount of boxes. Each of the following 2N - 1 lines contains numbers ai and oi —
amount of apples and oranges in the i-th box (0 ≤ ai, oi ≤ 109).
The sum of N in all the tests in the input doesn't exceed 105. All the input numbers are integer.

Output

For each test output two lines. In the first line output YES, if it's possible to choose N boxes, or NO otherwise.
If the answer is positive output in the second line N numbers — indexes of the chosen boxes. Boxes are numbered from 1 in the input order. Otherwise leave the second line empty. Separate the
numbers with one space.

Sample Input

Input

2
2
10 15
5 7
20 18
1
0 0

Output

YES
1 3
YES
1

#include<bits/stdc++.h>
using namespace std;
const int N = 2*1e5+10;
struct node
{
int x;
int y;
int index;
}a
;

bool cmp(node t1,node t2)
{
if(t1.x<t2.x) return true;
return false;
}
int main()
{
int T,i,j,n;
scanf("%d",&T);
while(T--) {
scanf("%d",&n);
for(i=1;i<=n+n-1;i++) {
scanf("%d%d",&a[i].x,&a[i].y);
a[i].index=i;
}
sort(a+1,a+1+n+n-1,cmp);
printf("YES\n");
for(i=2;i<n+n-1;i=i+2) {
if(a[i].y>a[i-1].y) printf("%d ",a[i].index);
else printf("%d ",a[i-1].index);
}
printf("%d\n",a[n+n-1].index);
}
return 0;
}