CCPC2018-湖南全国邀请赛-重现赛F题sorting

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原创最新推荐文章于 2025-09-23 10:09:09 发布 763 阅读 AI 写同款· GEO 优化 ›

最小路径和——动态规划求解（Java实现）最小路径和——动态规划求解（Java实现）题目：给定一个包含非负整数的 m x n 网格，请找出一条从左上角到右下角的路径，使得路径上的数字总和为最小。说明：每次只能向下或者向右移动一步。示例: 输入: [ [1,3,1], [1,5,1], [4,2,1] ] 输出: 7 解释: 因为路径 1→3→1→1→1 的总和最小。这题和机器人走路的思路类似，就是状态方程有所不同，... 阅读详情

Sorting

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 0 Accepted Submission(s): 0

Problem Description

Bobo has

n tuples

(a1,b1,c1),(a2,b2,c2),…,(an,bn,cn).
He would like to find the lexicographically smallest permutation

p1,p2,…,pn of

1,2,…,n such that for

i∈{2,3,…,n} it holds that

a p i - 1 + b p i - 1 a p i - 1 + b p i - 1 + c p i - 1 \leq a p i + b p i a p i + b p i + c p i .

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

ai,

bi and

ci.

Output

For each test case, print

n integers

p1,p2,…,pn seperated by spaces.
DO NOT print trailing spaces.

## Constraint

*

1≤n≤103
*

1≤ai,bi,ci≤2×109
* The sum of

n does not exceed

104.

Sample Input

Sample Output

2 1 1 2

1 2 3

题解：相邻两组的a[i-1],b[i-1],c[i-1]和a[i],b[i],c[i]满足上面的公式，如果相等的话就不交换两组的顺序。

代码：一开始我把大于和小于两种情况一起考虑，在排序的时候把相等的交换了位置，然后一直在wa。。。，果然还是太菜了。。

#include <iostream>
#include <cstdio>
#include <string>
#include <algorithm>
#include <cstdlib>

using namespace std;

const int MAX=10010;

struct node{
    long double x,y,z;
}a[MAX];
int b[MAX];

int main(){
    int n;
    while(cin >> n){
        for(int i = 1;i<=n;++i){
            cin >> a[i].x >> a[i].y >> a[i].z;
            b[i] = i;
        }
        for(int i = 1;i<=n;++i){
            for(int j = i+1;j<=n;++j){
                if(a[j].z*(a[i].x+a[i].y)>a[i].z*(a[j].x+a[j].y)){
                    swap(b[i],b[j]);
                    swap(a[i],a[j]); 
                }
                else if(a[j].z*(a[i].x+a[i].y)==a[i].z*(a[j].x+a[j].y) && b[i]>b[j]){
                    swap(b[i],b[j]);
                    swap(a[i],a[j]);
                }
        }
    }
        for(int i = 1;i<=n;++i){
            printf(i==1?"%d":" %d",b[i]);
        }
        cout << endl;
    }
    return 0;
}