HDU 2115 I Love This Game（结构体排序 or pair）

I Love This Game

Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 7163 Accepted Submission(s): 2451



[align=left]Problem Description[/align]
Do you like playing basketball ? If you are , you may know the NBA Skills Challenge . It is the content of the basketball skills . It include several parts , such as passing , shooting , and so on. After
completion of the content , the player who takes the shortest time will be the winner . Now give you their names and the time of finishing the competition , your task is to give out the rank of them ; please output their name and the rank, if they have the
same time , the rank of them will be the same ,but you should output their names in lexicographic order.You may assume the names of the players are unique.

Is it a very simple problem for you? Please accept it in ten minutes.

[align=left]Input[/align]
This problem contains multiple test cases! Ease test case contain a n(1<=n<=10) shows the number of players,then n lines will be given. Each line will contain the name of player and the time(mm:ss) of
their finish.The end of the input will be indicated by an integer value of zero.

[align=left]Output[/align]
The output format is shown as sample below.

Please output the rank of all players, the output format is shown as sample below;

Output a blank line between two cases.

[align=left]Sample Input[/align]

10
Iverson 17:19
Bryant 07:03
Nash 09:33
Wade 07:03
Davies 11:13
Carter 14:28
Jordan 29:34
James 20:48
Parker 24:49
Kidd 26:46
0

[align=left]Sample Output[/align]

Case #1
Bryant 1
Wade 1
Nash 3
Davies 4
Carter 5
Iverson 6
James 7
Parker 8
Kidd 9
Jordan 10

[align=left]Author[/align]
為傑沉倫

[align=left]Source[/align]
HDU 2007-10 Programming Contest_WarmUp

题解：结构体 or pair<string,string>e[12].

AC代码：

#include<iostream>
#include<memory.h>
#include<cstdlib>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<string>
#include<cstdlib>
#include<iomanip>
#include<vector>
#include<list>
#include<map>
#include<queue>
#include<algorithm>
typedef long long LL;
using namespace std;
//数学公式：1^3+2^3+3^3+....n^3=[n(n+1)/2]^2
struct node
{
char name[10];
int m,s;
}p[11];
bool cmp(node a,node b)
{
if(a.m==b.m&&a.s==b.s)
return strcmp(a.name,b.name)<0; //按字典序从小到大
else if(a.m==b.m)
return a.s<b.s; //从小到大
return a.m<b.m;
}
int main()
{
bool ok=0;
int i,j,n,ca=1,r,sum;
while(~scanf("%d",&n),n)
{
if(ok)
printf("\n");
ok=1;
for(i=0;i<n;i++)
{
scanf("%s %d:%d",p[i].name,&p[i].m,&p[i].s);
}
r=1,sum=1;
sort(p,p+n,cmp);
printf("Case #%d\n",ca++);
for(i=0;i<n;i++)
{
if(i!=0&&(p[i].m!=p[i-1].m||p[i].s!=p[i-1].s))
r=sum;   //r为名次
printf("%s %d\n",p[i].name,r);
sum++;
}
}
return 0;
}

AC：pair

#include <iostream>
#include <utility>
#include <algorithm>
#include <cstring>
#include <cmath>

using namespace std;
//#define LOCAL

int main()
{
#ifdef LOCAL
freopen("in.txt","r",stdin);
#endif
int T=0,n,mc[12];
string str1,str2;
pair<string,string> e[12];
while(cin>>n,n)
{
if(T) cout<<endl;
cout<<"Case #"<<++T<<endl;
if(T>1) cout<<endl;
for(int i=0;i<n;++i)
{
cin>>str1>>str2;
e[i]=make_pair(str2,str1);
}
sort(e,e+n);
mc[0]=1;
for(int i=1;i<n;++i)
if(e[i].first==e[i-1].first)
mc[i]=mc[i-1];
else
mc[i]=i+1;
for(int i=0;i<n;++i)
cout<<e[i].second<<' '<<mc[i]<<endl;
}
return 0;
}