HDU：5499 SDOI（结构体技巧）

SDOI

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 896    Accepted Submission(s): 358

Problem Description

The Annual National Olympic of Information(NOI) will be held.The province of Shandong hold a Select(which we call SDOI for short) to choose some people to go to the NOI. n(n≤100) people comes to the Select and there is m(m≤50) people who can go to the NOI.

According to the tradition and regulation.There were two rounds of the SDOI, they are so called "Round 1" and "Round 2", the full marks of each round is 300.

All the n people take part in Round1 and Round2, now the original mark of every person is known. The rule of SDOI of ranking gets to the "standard mark". For each round there is a highest original mark,let's assume that is x.(it is promised that not all person in one round is 0,in another way,x>0). So for this round,everyone's final mark equals to his/her original mark∗(300/x).

After we got everyone's final mark in both round.We calculate the Ultimate mark of everyone as 0.3∗round1′s final mark + 0.7∗round2′s final mark.It is so great that there were no two persons who have the same Ultimate mark.

After we got everyone's Ultimate mark.We choose the persons as followed:

To encourage girls to take part in the Olympic of Information.In each province,there has to be a girl in its teams.

1. If there is no girls take part in SDOI,The boys with the rank of first m enter the team.
2. If there is girls, then the girl who had the highest score(compared with other girls) enter the team,and other(boys and other girls) m-1 people with the highest mark enter the team.

Just now all the examination had been finished.Please write a program, according to the input information of every people(Name, Sex ,The original mark of Round1 and Round2),Output the List of who can enter the team with their Ultimate mark decreasing.

 

Input

There is an integer T(T≤100) in the first line for the number of testcases and followed T testcases.

For each testcase, there are two integers n and m in the first line(n≥m), standing for the number of people take part in SDOI and the allowance of the team.Followed with n lines,each line is an information of a person. Name(A string with length less than 20,only contain numbers and English letters),Sex(male or female),the Original mark of Round1 and Round2 (both equal to or less than 300) separated with a space.

 

Output

For each testcase, output "The member list of Shandong team is as follows:" without Quotation marks.

Followed m lines,every line is the name of the team with their Ultimate mark decreasing.

 

Sample Input

210 8dxy male 230 225davidwang male 218 235evensgn male 150 175tpkuangmo female 34 21guncuye male 5 15faebdc male 245 250lavender female 220 216qmqmqm male 250 245davidlee male 240 160dxymeizi female 205 1902 1dxy male 300 300dxymeizi female 0 0

 

Sample Output

The member list of Shandong team is as follows:faebdcqmqmqmdavidwangdxylavenderdxymeizidavidleeevensgnThe member list of Shandong team is as follows:dxymeiziHintFor the first testcase: the highest mark of Round1 if 250,so every one's mark times(300/250)=1.2, it's same to Round2.The Final of The Ultimate score is as followedfaebdc 298.20qmqmqm 295.80davidwang 275.88dxy 271.80lavender 260.64dxymeizi 233.40davidlee 220.80evensgn 201.00tpkuangmo 29.88guncuye 14.40For the second testcase,There is a girl and the girl with the highest mark dxymeizi enter the team, dxy who with the highest mark,poorly,can not enter the team.

 

Source

BestCoder Round #59 (div.2)

 

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hujie

题目大意：选一群人去参加比赛，所以经过热身赛选拔，在这个热身赛中，如果没有女参加的话，就取最终成绩的前m名去参加比赛，如果有女的参加的话，最终选拔出来的队伍里必须最低有一个女性，即使按正常排位前m名没有女性，那么也要抽出后面的一位女性来替换第m名进入最终队伍；如果正常排位里前m名有女性，则按排名输出。

比赛的最终成绩=(第一圈的最高分/300)*第一圈的分数*0.3+(第二圈的最高分/300)*第二圈的分数*0.7

代码如下：

#include <stdio.h>#include <string.h>#include <algorithm>using namespace std;struct node{int sex;//sex=0为女性，sex=1为男性 char name[25];double score;double round1;double round2;int id;//没用到 }per[110];bool cmp(struct node a,struct node b){return a.score>b.score;}int main(){int t;scanf("%d",&t);while(t--){int n,m;scanf("%d%d",&n,&m);double max1=-1,max2=-1;//求第一圈的最高分和第二圈的最高分 int flagnv=0;//判断是否有女性参加比赛 for(int i=0;i<n;i++){char xing[10];scanf("%s %s %lf %lf",&per[i].name,xing,&per[i].round1,&per[i].round2);per[i].id=i;if(strcmp(xing,"male")==0)per[i].sex=1;else{per[i].sex=0;flagnv=1;}if(per[i].round1>max1){max1=per[i].round1;}if(per[i].round2>max2){max2=per[i].round2;}}for(int i=0;i<n;i++){per[i].score=(300/max1)*per[i].round1*0.3+(300/max2)*per[i].round2*0.7;//算分 }sort(per,per+n,cmp);if(flagnv==0)//没女生，直接按排名输出 {printf("The member list of Shandong team is as follows:\n");for(int i=0;i<m;i++){printf("%s\n",per[i].name);}}else{int flag;//标记排名中第一位女生排第几位 for(int i=0;i<n;i++){if(per[i].sex==0){flag=i;break;}}if(flag>m-1)//第一位女生不在选拔的队伍里，即选出的队伍里没有女生，那么让她替代队伍里的最后一名 {strcpy(per[m-1].name,per[flag].name);printf("The member list of Shandong team is as follows:\n");for(int i=0;i<m;i++){printf("%s\n",per[i].name);}}else//选拔出来的队伍里有女生，那么直接按排名输出 {printf("The member list of Shandong team is as follows:\n");for(int i=0;i<m;i++){printf("%s\n",per[i].name);}}}}return 0;}