CodeForces 23C-Oranges and Apples

解决一个算法挑战,从2N-1个盒子中选择N个盒子,确保包含至少一半的苹果和橙子。输入包括测试数量及每个盒子的苹果和橙子数量。

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Oranges and Apples
time limit per test
 2 seconds
memory limit per test
 256 megabytes
input
 standard input
output
 standard output

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 NOotherwise. 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.

Examples
input
2
2
10 15
5 7
20 18
1
0 0
output
YES
1 3
YES
1

#include <iostream>
#include <algorithm>
#include <stdio.h>

using namespace std;

struct node
{
    int a,b,o;
} x[200009];

int f(node p,node q)
{
    return p.a<q.a;
}

int main()
{
    int t,n,N,i;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        N=2*n-1;
        for(i=1; i<=N; i++)
        {
            scanf("%d%d",&x[i].a,&x[i].b);
            x[i].o=i;
        }
        sort(x+1,x+N+1,f);
        printf("YES\n");
        for(i=2; i<N; i=i+2)
        {
            if(x[i].b>x[i-1].b)
                printf("%d ",x[i].o);
            else
                printf("%d ",x[i-1].o);
        }
        printf("%d\n",x[N].o);
    }
    return 0;
}

### Codeforces Problem 976C Solution in Python For solving problem 976C on Codeforces using Python, efficiency becomes a critical factor due to strict time limits aimed at distinguishing between efficient and less efficient solutions[^1]. Given these constraints, it is advisable to focus on optimizing algorithms and choosing appropriate data structures. The provided code snippet offers insight into handling string manipulation problems efficiently by customizing comparison logic for sorting elements based on specific criteria[^2]. However, for addressing problem 976C specifically, which involves determining the winner ('A' or 'B') based on frequency counts within given inputs, one can adapt similar principles of optimization but tailored towards counting occurrences directly as shown below: ```python from collections import Counter def determine_winner(): for _ in range(int(input())): count_map = Counter(input().strip()) result = "A" if count_map['A'] > count_map['B'] else "B" print(result) determine_winner() ``` This approach leverages `Counter` from Python’s built-in `collections` module to quickly tally up instances of 'A' versus 'B'. By iterating over multiple test cases through a loop defined by user input, this method ensures that comparisons are made accurately while maintaining performance standards required under tight computational resources[^3]. To further enhance execution speed when working with Python, consider submitting codes via platforms like PyPy instead of traditional interpreters whenever possible since they offer better runtime efficiencies especially important during competitive programming contests where milliseconds matter significantly.
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