山东省第二届ACM大学生程序设计竞赛 The Android University ACM Team Selection Contest
The Android University ACM Team Selection Contest
Time Limit: 1000MS Memory limit: 65536K
题目描述
To be selected, one team has to solve at least one problem in the contest. The the top M teams who solved at least one problem are selected (If there are less than M teams solving at least one problem, they are all selected).
There is an bonus for the girls - if top M teams contains no all-girls teams,the highest ranked all-girls team is also selected (together with the M top teams), provided that they have solved at least one problem.
Recall that in an ACM/ICPC style contest, teams are ranked as following:
1. The more problems a team solves, the higher order it has.
2. If multiple teams have the same number of solved problems, a team with a smaller penalty value has a higher order than a team with a
larger penalty value.
Given the number of teams N, the number M defined above, and each team's name, number of solved problems, penalty value and whether it's an all-girls team, you are required to write a program to find out which teams are selected.
输入
Each test case begins with a line contains two integers, N (1 <= N <=10^4) and M (1 <= M <= N), separated by a single space. Next will be N lines, each of which gives the information about one specific competing team.Each of the N lines contains a string S (with length at most 30, and consists of upper and lower case alphabetic characters) followed by three integers, A(0 <= A <= 10), T (0 <= T <= 10) and P (0 <= P <= 5000), where S is the name of the team, A indicates whether the team is an all-girls team (it is not an all-girls team if Ai is 0, otherwise it is an all-girls team). T is the number of problems the team solved, and P is the penalty value of the team.
The input guarantees that no two teams who solved at least one problem have both the same T and P.
输出
示例输入
3 5 3 AU001 0 0 0 AU002 1 1 200 AU003 1 1 30 AU004 0 5 500 AU005 0 7 1000 2 1 BOYS 0 10 1200 GIRLS 10 1 290 3 3 red 0 0 0 green 0 0 0 blue 0 1 30
示例输出
Case 1: AU003 AU004 AU005 Case 2: BOYS GIRLS Case 3: blue 3 3
提示
来源
示例程序
坑点是输出排名的时候是按照输入时候的顺序 英语差就是坑
ACcode：
#include <map>
#include <queue>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <stdlib.h>
#include <iostream>
#include <algorithm>
#define maxn 10010
using namespace std;
struct Node{
char s[100];
bool flag;
int num;
int sorce;
int id;
void in(){
scanf("%s%d%d%d",&s,&flag,&num,&sorce);
}
void on(){
printf("%s\n",s);
}
bool operator<(const Node &a)const{
if(a.num==this->num)return this->sorce<a.sorce;
return this->num>a.num;
}
}my[maxn];
bool cmp(Node a,Node b){return a.id<b.id;}
int n,m,loop,t=1,cnt;
void doit(){
scanf("%d%d",&n,&m);
bool is=false;
Node temp;
cnt=0;
memset(my,0,sizeof(my));
for(int i=0;i<n;++i){
temp.in();
if(temp.num){
my[cnt]=temp;
my[cnt].id=cnt;
if(my[cnt].flag)is=true;
cnt++;
}
}
printf("Case %d:\n",t++);
sort(my,my+cnt);
int numm=m,mm=m;
bool flag=true;
sort(my,my+m,cmp);
for(int i=0;i<m;++i)
if(my[i].num){
if(my[i].flag)
flag=false;
my[i].on();
numm--;
}
else break;
if(flag&&is)
for(int i=mm-1;i<cnt;++i)
if(my[i].num&&my[i].flag){
my[i].on();
numm--;
break;
}
while(numm>0){
printf("%d\n",mm);
numm--;
}
printf("\n");
}
int main(){
scanf("%d",&loop);
while(loop--)doit();
return 0;
}
/*
3
5 3
AU001 0 0 0
AU002 1 1 200
AU003 1 1 30
AU004 0 5 500
AU005 0 7 1000
2 1
BOYS 0 10 1200
GIRLS 10 1 290
3 3
red 0 0 0
green 0 0 0
blue 0 1 30
*/