CF23C 思维

最新推荐文章于 2022-10-15 14:46:08 发布

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最新推荐文章于 2022-10-15 14:46:08 发布

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http://codeforces.com/problemset/problem/23/C

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 test(s) 
      

          input 
        
2
2
10 15
5 7
20 18
1
0 0

          output 
        
YES
1 3
YES
1
 
         
 
        

/**
CF23C 思维
题目大意：给定2*n-1个箱子，里面有ai的苹果，oi的橘子，问怎样选出n个使得橘子苹果的数量和都不小于总数的一半
解题思路：一开始按照x排序，那么从取到的第2个箱子开始，不会比上一次舍去的箱子里的x的数量少，再加上取得第一个箱子里面的x
         说明x肯定比总数的一半多；y的话，每次两个里面取走一个大的，取得总数肯定不会比舍去的少。
         这样取保证有解。
*/
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <iostream>
using namespace std;
const int maxn=200004;
struct note
{
    int x,y,ip;
}a[maxn];

bool cmp(note a,note b)
{
    if(a.x==b.x)
        return a.y<b.y;
    return a.x<b.x;
}

int n,ans[maxn];

int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&n);
        for(int i=1;i<=2*n-1;i++)
        {
            scanf("%d%d",&a[i].x,&a[i].y);
            a[i].ip=i;
        }
        sort(a+1,a+n*2,cmp);
        int k=0;
        for(int i=1;i<2*n-1;i+=2)
        {
            if(a[i].y>=a[i+1].y)
                ans[k++]=a[i].ip;
            else
                ans[k++]=a[i+1].ip;
        }
        ans[k++]=a[2*n-1].ip;
        puts("YES");
        sort(ans,ans+k);
        for(int i=0;i<k;i++)
        printf(i==k-1?"%d\n":"%d ",ans[i]);
    }
    return 0;
}