sort(HDU-3293)

Problem Description

As is known to all, long long ago sailormoon once was an association of fighters. Till now, sailormoon is also an association of girls. Owe to some unknown reasons, girls are necessary to fight for peace.
Their boss, lcy, wants to strengthen their ability, so he give them his precious collections---weapons for many years. Because these collections are really age-old, it is hard to recognize from one to another. So girls intend to sort them before they use. Each weapon has its name, origin and level of harmfulness ( level contains three ranks: wonderful, good, so-so).
In order to make it clear, girls want to sort like this:
firstly,sort according to the origin (sort by lexicographic order), if two or more have the same origin, they will be sorted together;
secondly, sort according ranks, wonderful is the best, good is next, the third is so-so;
thirdly, if two or more have same origin and rank, sort them according to the lexicographic order.

 

Input

Input contains multiply cases. Each case contains several lines. First line is an integer N(0<N<=500), representing the number of weapons. Then N lines follows. Each line represent a kind of weapon, and contains a set of strings representing name, origin and level of harmfulness. 
Each string will not exceed 20 characters.
Sure that same origin will not exist the same weapon.

 

Output

Please output your list after sorting (format according to sample, pay attention to the spaces,ten spaces need ^ ^).

 

Sample Input

5
knife qizhou so-so
gun qizhou wonderful
knife zhengzhou good
stick zhengzhou good
rope shengzhou so-so

 

Sample Output

Case 1
qizhou:
          gun wonderful
          knife so-so
shengzhou:
          rope so-so
zhengzhou:
          knife good
          stick good

 

代码如下：

#include<bits/stdc++.h>
using namespace std;
struct pp
{
    string ori,na,le;
    int l;
};
pp p[510];
bool cmp(const pp&a,const pp&b)
{
    if(a.ori!=b.ori)
        return a.ori<b.ori;
    else if(a.l!=b.l)
        return a.l<b.l;
    else
        return a.na<b.na;
}
set<string>origin;
int main()
{
    int n,t=1;
    while(cin>>n)
    {
        string a,b,c;
        for(int i=0;i<n;i++)
        {
            cin>>p[i].na>>p[i].ori>>p[i].le;
            origin.insert(p[i].ori);
            if(p[i].le[0]=='w')
                p[i].l=1;
            else if(p[i].le[0]=='g')
                p[i].l=2;
            else
                p[i].l=3;
        }
        int l=0;
        sort(p,p+n,cmp);
        cout<<"Case "<<t<<endl;
        for(set<string>::iterator it=origin.begin();it!=origin.end();it++)
        {
            cout<<*it<<":\n";
            for(;l<n&&*it==p[l].ori;l++)
                cout<<"          "<<p[l].na<<" "<<p[l].le<<endl;
        }
        t++;
        origin.clear();
    }
    return 0;
}