HDU 5182 PM2.5(排序)——BestCoder Round #32

PM2.5

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

Problem Description

Nowadays we use content of PM2.5 to discribe the quality of air. The lower content of PM2.5 one city have, the better quality of air it have. So we sort the cities according to the content of PM2.5 in asending order.

Sometimes one city’s rank may raise, however the content of PM2.5 of this city may raise too. It means that the quality of this city is not promoted. So this way of sort is not rational. In order to make it reasonable, we come up with a new way to sort the cityes. We order this cities through the diffrence between twice measurement of content of PM2.5 (first measurement – second measurement) in descending order, if there are ties, order them by the second measurement in asending order , if also tie, order them according to the input order.

 

Input

Multi test cases (about  100 ), every case contains an integer  n  which represents there are  n  cities to be sorted in the first line.
Cities are numbered through 0 to  n−1 .
In the next  n  lines each line contains two integers which represent the first and second measurement of content of PM2.5
The ith line describes the information of city  i−1
Please process to the end of file.

[Technical Specification]
all integers are in the range  [1,100]

 

Output

For each case, output the cities’ id in one line according to their order.

 

Sample Input

   
   
   
   

    
    
    
    2
100 1
1 2
3
100 50
3 4
1 2
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    0 1
0 2 1
   
   
   
   

 

Source

BestCoder Round #32

 

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出题人的解题思路：

对于输入的每一行一两个整数作差，按照差值从大到小排序，如果差值一样，按照后面的整数从小到大排序，如果还是一样按照ID从小到大排序。

题意：给你0~n-1这n个城市两次测量的PM2.5值，要求按照规定方式进行排序后，输出城市的编号

排序要求：从①到③，优先级递减

①第一次测量的PM2.5值-第二次测量的PM2.5值 差值大的排前面

②第二次测量的PM2.5值大的排前面

③输入早的排前面，即编号小的排前面

简单的排序题，在此不做过多说明，详细见代码

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<queue>
#include<math.h>
#include<vector>
#include<map>
#include<set>
#include<stdlib.h>
#include<cmath>
#include<string>
#include<algorithm>
#include<iostream>
#define exp 1e-10
using namespace std;
const int N = 105;
const int inf = 1000000000;
const int mod = 258280327;
struct city
{
    int f,s,c,id;
}w[N];
bool cmp(city x,city y)
{
    if(x.c!=y.c)
        return x.c>y.c;
    if(x.s!=y.s)
        return x.s<y.s;
    return x.id<y.id;
}
int main()
{
    int n,i;
    while(~scanf("%d",&n))
    {
        for(i=0;i<n;i++)
        {
            scanf("%d%d",&w[i].f,&w[i].s);
            w[i].c=w[i].f-w[i].s;
            w[i].id=i;
        }
        sort(w,w+n,cmp);
        for(i=0;i<n-1;i++)
            printf("%d ",w[i].id);
        printf("%d\n",w[i].id);
    }
    return 0;
}

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