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

题意

简单来说，每个城市都有两个PM值，排序的时候让你按照两个PM值之差，从大到小排序，然后再按照第二个PM值，从小到大，再按照id从小到大排序

题解

照着题意写cmp，然后调用stl的sort就好～

struct node
{
    int x;
    int y;
    int ss;
    int id;
};
bool cmp(node a,node b)
{
    if(a.ss==b.ss&&a.y==b.y)
        return a.id<b.id;
    if(a.ss==b.ss)
        return a.y<b.y;
    return a.ss>b.ss;
}
node kiss[maxn];
int main()
{
    int n;
    while(cin>>n)
    {
        REP(i,n)
        {
            RD(kiss[i].x);
            RD(kiss[i].y);
            kiss[i].ss=kiss[i].x-kiss[i].y;
            kiss[i].id=i;
        }
        sort(kiss,kiss+n,cmp);
        int flag=1;
        REP(i,n)
        {
            if(flag)
            {
                cout<<kiss[i].id;
                flag=0;
            }
            else
                cout<<" "<<kiss[i].id;
        }
        cout<<endl;
    }
}