题目链接：
PKU:http://poj.org/problem?id=3339
HDU:http://acm.hdu.edu.cn/showproblem.php?pid=2409
Description
Barry Bennett, the coach of the Bings football team, wants to arrange his team for an important match against the Bangs. He decides on the formation he wants to play, for example 4-4-2, meaning that there will be four defenders, four midfielders, and two strikers (and of course, one goalkeeper). Your task is to determine the players who will play. For each available player, we know his role (e.g. midfielder). For each role, the players are selected in ascending order of their numbers. When the players are selected, you must determine the captain too, who is the player with the longest record in the team play. In case two players have the same record, the one with bigger number is chosen. Note that the captain is chosen among the players that are selected in the arrange.
Input
The input consists of multiple test cases. The first 22 lines of each test case contain the data for the 22 available players in this format:
number name role year1–year‘1 year2–year‘2 ...
number is the player number (unique positive integer less than 100). name is a string of at most 20 letters. role is a single character among G, D, M, S, for goalkeeper, defender, midfielder, and striker respectively. Each yeari –year‘i pair (yeari ≤ year‘i) shows the player has been a member of the team between the specified years (inclusive). The years are in four-digit format. There is at least one and at most 20 such pairs, and the same year is not repeated more than once in the list. There is a 23rd line describing the desired formation, like 4-4-2 in that format. Note that there are only three numbers in the formation (so, 4-3-2-1 is not valid), none of them is zero, and their sum is always 10. The input is terminated by a line containing a single 0.
Output
For each test case, output a list of 11 players chosen in the arrange. Each line must contain the player number, his name and his role, separated by single blank characters. The players must be sorted according to their role, in the order of goalkeeper,
defenders, midfielders, and strikers. The players with the same role are sorted according to ascending order of their numbers. There is an exception that the captain always comes as the first player in the entire list. If it is not possible to arrange the
team conforming to the desired formation, write a single line containing IMPOSSIBLE TO ARRANGE in the output. There should be a blank line after the output for each test case.
Sample Input
9 PlayerA M 2000-2001 2003-2006 2 PlayerB M 2004-2006 10 PlayerC D 2001-2005 1 PlayerD D 2000-2001 2002-2004 11 PlayerE S 2003-2006 8 PlayerF M 2005-2006 22 PlayerG S 2005-2006 25 PlayerH G 2000-2001 2002-2003 2005-2006 6 PlayerI D 2003-2006 26 PlayerJ D 2003-2004 2000-2001 18 PlayerK M 2003-2004 19 PlayerL M 2000-2001 2003-2006 7 PlayerM S 2003-2006 1999-2001 21 PlayerN S 2003-2006 13 PlayerO S 2005-2006 15 PlayerP G 2001-2006 14 PlayerQ D 2003-2004 5 PlayerR S 2000-2005 20 PlayerS G 2000-2002 2003-2003 12 PlayerT M 2004-2005 3 PlayerU D 2000-2005 4 PlayerV M 2001-2004 4-4-2 0
Sample Output
7 PlayerM S 15 PlayerP G 1 PlayerD D 3 PlayerU D 6 PlayerI D 10 PlayerC D 2 PlayerB M 4 PlayerV M 8 PlayerF M 9 PlayerA M 5 PlayerR S
Source
题意：
给出22个球员的各种信息，要求按照给出的阵形选择球员和队长，给出的阵形连守门员一共十一人；
选择球员的规则是：同角色的球员按照编号从小到大选择，直到选够此角色的人数；
选择队长的规则是：在所有已经被选择为某角色的球员中，选择服役时间最长的那个球员，
如果服役时间时间相同，就选择编号较大的那一个球员；
PS:此题有点麻烦！
代码如下：
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
using namespace std;
struct ROLE
{
int num;
char name[47];
char R;
int time;
int flag;
} p[47];
bool cmp1(ROLE a, ROLE b)
{
return a.num < b.num;
}
bool cmp2(ROLE a, ROLE b)
{
return a.flag < b.flag;
}
int main()
{
char s[700];
while(scanf("%d",&p[0].num)!=EOF)
{
if(p[0].num == 0)
break;
for(int i = 0; i < 27; i++)
p[i].flag = 0;
for(int i = 0; i < 22; i++)
{
memset(s,'\0',sizeof(s));
int sum = 0;
if(i > 0)
scanf("%d",&p[i].num);
scanf("%s",p[i].name);
getchar();
scanf("%c",&p[i].R);
gets(s);
for(int j = 0; j < strlen(s)-4; j++)
{
if(s[j]==' ')
continue;
int t1 = 0, t2 = 0;
for(int l = j; l < j+4; l++)
{
t1=t1*10+s[l]-'0';
}
j+=4;
if(s[j] == '-')
{
for(int l = j+1; l < j+5; l++)
{
t2=t2*10+s[l]-'0';
}
}
j+=5;
sum+=t2-t1+1;
}
p[i].time = sum;
//printf("%d....\n",sum);
}
sort(p,p+22,cmp1);
char ss[17];
scanf("%s",ss);
int a, b, c;
a = ss[0]-'0';
b = ss[2]-'0';
c = ss[4]-'0';
//printf("%d %d %d\n",a,b,c);
int l = 1;
for(int i = 0; i < 22; i++)
{
if(p[i].R == 'G')
{
p[i].flag = l;
l++;
break;
}
}
for(int i = 0; i < a; i++)
{
for(int j = 0; j < 22; j++)
{
if(p[j].R == 'D' && p[j].flag==0)
{
p[j].flag = l;
l++;
break;
}
}
}
for(int i = 0; i < b; i++)
{
for(int j = 0; j < 22; j++)
{
if(p[j].R == 'M' && p[j].flag==0)
{
p[j].flag = l;
l++;
break;
}
}
}
for(int i = 0; i < c; i++)
{
for(int j = 0; j < 22; j++)
{
if(p[j].R == 'S' && p[j].flag==0)
{
p[j].flag = l;
l++;
break;
}
}
}
if(l!=12 ||(a+b+c!=10))
{
printf("IMPOSSIBLE TO ARRANGE\n\n");
continue;
}
sort(p,p+22,cmp2);
int maxx = -1;
int pos = 0;
for(int i = 0; i < 22; i++)
{
if(p[i].flag && p[i].time > maxx)
{
maxx = p[i].time;
pos = i;
}
else if(p[i].flag && p[i].time == maxx)
{
if(p[pos].num < p[i].num)
pos = i;
}
}
printf("%d %s %c\n",p[pos].num,p[pos].name,p[pos].R);
for(int i = 0; i < 22; i++)
{
if(p[i].flag && i!=pos)
printf("%d %s %c\n",p[i].num,p[i].name,p[i].R);
}
printf("\n");
}
return 0;
}
POJ 3339 & HDU 2409 Team Arrangement(结构体)
原文地址：http://blog.csdn.net/u012860063/article/details/39757219
