HDOJ 5499 SDOI 【结构体排序】
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SDOI
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 690 Accepted Submission(s): 271
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.
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
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
After we got everyone's final mark in both round.We calculate the Ultimate mark of everyone as
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 integersn 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.
For each testcase, there are two integers
Output
For each testcase, output "The member list of Shandong team is as follows:" without Quotation marks.
Followedm lines,every line is the name of the team with their Ultimate mark decreasing.
Followed
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.
恩,题目大意就是说,两轮比赛,没轮比赛中,找出最大值,将最大值换算成满分300分要乘以 p 然后让每个人的成绩都乘以 p 以至所有人的成绩不同,然后第一轮成绩换算过后乘以0.3加上第二轮成绩换算过后乘以0.7就是最后成绩。然后在有女生参赛的情况下,所选出的 m 个人中必须有一个女生,剩下 m-1 个人按成绩从高到低,没有女生参赛的话就直接从高到低 m 个人。
<span style="font-family:Comic Sans MS;">#include <iostream>#include<cstdio>#include<cstring>#include<algorithm>#define maxn 110using namespace std;struct node{ char name[30]; int s; double mark;}peo[maxn];int cmp(node a,node b){ return a.mark>b.mark;}int main(){ int t,n,m; int mark1[maxn],mark2[maxn]; double p1,p2; char sex[10]; scanf("%d",&t); while(t--) { int sign=0; int m1=-1; int m2=-1; scanf("%d%d",&n,&m); for(int i=0;i<n;++i) { scanf("%s%s",&peo[i].name,sex); if(sex[0]=='m') peo[i].s=1; else { sign=1; peo[i].s=0; } scanf("%d%d",&mark1[i],&mark2[i]); m1=max(m1,mark1[i]); m2=max(m2,mark2[i]); } p1=300*1.0/m1; p2=300*1.0/m2; for(int i=0;i<n;++i) peo[i].mark=mark1[i]*p1*0.3+mark2[i]*p2*0.7; sort(peo,peo+n,cmp); printf("The member list of Shandong team is as follows:\n"); if(!sign) { for(int i=0;i<m;++i) printf("%s\n",peo[i].name); } else { int cnt=0,ms=0; for(int i=0;i<n;++i) { cnt++; if(cnt<m) { if(!peo[i].s) ms=1; printf("%s\n",peo[i].name); } else { if(ms) { printf("%s\n",peo[i].name); break; } else { if(!peo[i].s) { printf("%s\n",peo[i].name); break; } } } } } } return 0;}</span>
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