2018湘潭邀请赛F题题解_2018ti杯f题-优快云博客

本文链接：https://blog.youkuaiyun.com/songziqi98/article/details/80343203

博客详细解析了2018湘潭邀请赛中F题的解题思路，题目要求找到最小字典序的排列，使得对于i从2到n，条件ai * pi < ai * pi-1成立。博主指出在实现时需要注意数值精度问题，提供了通过测试的AC代码。

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2018湘潭邀请赛F题题解

F. Sorting

Bobo has n tuples (a₁, b₁, c₁), (a₂, b₂, c₂), . . . , (a_n, b_n, c_n). He would like to find the lexicographically smallest permutation p₁, p₂, . . . , p_n of 1, 2, . . . , n such that for i ∈ {2, 3, . . . , n} it holds that

Input

The input consists of several test cases and is terminated by end-of-file.

The first line of each test case contains an integer n. The i-th of the following n lines contains 3 integers a_i, b_i and c_i.

Output

For each test case, print n integers p₁, p₂, . . . , p_n seperated by spaces. DO NOT print trailing spaces.

Constraint

l The sum of n does not exceed 104.

l 1 ≤ n ≤ 10³

l 1 ≤ a_i, b_i, c_i ≤ 2*10⁹

Sample Input

1 1 1

1 1 2

1 1 1

1 3 1

2 2 1

3 1 1

Sample Output

2 1

1 2

1 2 3

相信大部分读完题的人第一反应是用sort的cmp函数比较，此题有个精度的坑。不过，比赛时队友运气好变量用double就过了，想向欧皇下跪，比赛结束感觉还是不对劲，遂按比赛思路再写一遍，果不其然wrong answer。

保证高精度也蛮简单，只要用无符号长整型存数据，然后把不等式交叉相乘再消下元即可。Too simple

附上AC代码：

#include <iostream>
#include <cstdio>
#include <algorithm>
using namespace std;

const int SIZE = 1e3+50;
struct node {
    unsigned long long a, b, c;
    int mark;
};
bool cmp(const node x, const node y) {
    auto m = (x.a+x.b)*(y.c);
    auto n = (x.c)*(y.a+y.b);
    if(m == n) return x.mark < y.mark;
    else return m < n;
}
int main()
{
    int n;
    while(scanf("%d", &n) != EOF) {
        node gru[SIZE];
        for(int i = 0; i < n; i++) {
            gru[i].mark = i+1;
            scanf("%llu %llu %llu", &gru[i].a, &gru[i].b, &gru[i].c);
        }
        sort(gru, gru+n, cmp);
        printf("%d", gru[0].mark);
        for(int i = 1; i < n; i++) {
            printf(" %d", gru[i].mark);
        }
        printf("\n");
    }
    return 0;
}