hdoj 5499 SDOI 【结构体排序】

SDOI

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 912    Accepted Submission(s): 365

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 is300.

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 isx.(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 as0.3∗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 withn lines,each line is an information of a person. Name(A string with length less than20,only contain numbers and English letters),Sex(male or female),the Original mark of Round1 and Round2 (both equal to or less than300) 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.

 

按题意先算出每个人的分数（在这儿出了小错），选拔要求：

1）没有女生，输出分数较高的m个人；

2）有女生，女生之中最高的一定输出，再输出其他男女生中m-1个分数较高的人

ps:

很多时候自己并不是不能做对，多一点自信，相信自己的思路是对的，可能就是某一处出了小问题，大的方向还是对的.

代码：

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;struct Node{char name[30];char sex[10];double r1,r2;double score; }stu[110]; int cmp1(Node a,Node b) { return a.r1>b.r1;  } int cmp2(Node a,Node b) { return a.r2>b.r2; } int cmp3(Node a,Node b) { return a.score >b.score;   } int main(){int t,n,m,i,j,f,f1,k;double m1,m2;char s[10]="female";scanf("%d",&t);while(t--){f=0;f1=0;scanf("%d%d",&n,&m);if(!m)continue; for(i=1;i<=n;i++){scanf("%s %s",stu[i].name,stu[i].sex);getchar();scanf("%lf%lf",&stu[i].r1,&stu[i].r2);}//两个cmp连用只有最后一个有效 sort(stu+1,stu+n+1,cmp1);m1=300.0/stu[1].r1;// 300/最大r1值sort(stu+1,stu+n+1,cmp2); m2=300.0/stu[1].r2;//300/最大r2值for(i=1;i<=n;i++)stu[i].score=0.3*m1*stu[i].r1+0.7*m2*stu[i].r2;sort(stu+1,stu+n+1,cmp3);printf("The member list of Shandong team is as follows:\n");//以下是选拔分析！！ for(i=1;i<=n;i++){if(strcmp(stu[i].sex,"female")==0){f=1;k=i;break;}}if(f){if(k<=m){for(i=1;i<=m;i++){printf("%s\n",stu[i].name);}}else{for(i=1;i<m;i++){printf("%s\n",stu[i].name); }printf("%s\n",stu[k].name);}}else{for(i=1;i<=m;i++){printf("%s\n",stu[i].name);}}}return 0;}