HDU 2115 I Love This Game（結構體排序 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
Problem Description 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.
Input 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.
Output 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.
Sample Input
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
Sample Output
Case #1
Bryant 1
Wade 1
Nash 3
Davies 4
Carter 5
Iverson 6
James 7
Parker 8
Kidd 9
Jordan 10
Author 為傑沉倫
Source HDU 2007-10 Programming Contest_WarmUp
題解：結構體 or pair<string,string>e[12].
AC代碼：
#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;
//數學公式：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; //按字典序從小到大
else if(a.m==b.m)
return a.s<b.s; //從小到大
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為名次
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;
}
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