1055. The World's Richest (25)－PAT
1055. The World's Richest (25)
Forbes magazine publishes every year its list of billionaires based on the annual ranking of the world's wealthiest people. Now you are supposed to simulate this job, but concentrate only on the people in a certain range of ages. That is, given the net worths of N people, you must find the M richest people in a given range of their ages.
Input Specification:
Each input file contains one test case. For each case, the first line contains 2 positive integers: N (<=105) - the total number of people, and K (<=103) - the number of queries. Then N lines follow, each contains the name (string of no more than 8 characters without space), age (integer in (0, 200]), and the net worth (integer in [-106, 106]) of a person. Finally there are K lines of queries, each contains three positive integers: M (<= 100) - the maximum number of outputs, and [Amin, Amax] which are the range of ages. All the numbers in a line are separated by a space.
Output Specification:
For each query, first print in a line "Case #X:" where X is the query number starting from 1. Then output the M richest people with their ages in the range [Amin, Amax]. Each person's information occupies a line, in the format
Name Age Net_WorthThe outputs must be in non-increasing order of the net worths. In case there are equal worths, it must be in non-decreasing order of the ages. If both worths and ages are the same, then the output must be in non-decreasing alphabetical order of the names. It is guaranteed that there is no two persons share all the same of the three pieces of information. In case no one is found, output "None".
Sample Input:
12 4 Zoe_Bill 35 2333 Bob_Volk 24 5888 Anny_Cin 95 999999 Williams 30 -22 Cindy 76 76000 Alice 18 88888 Joe_Mike 32 3222 Michael 5 300000 Rosemary 40 5888 Dobby 24 5888 Billy 24 5888 Nobody 5 0 4 15 45 4 30 35 4 5 95 1 45 50Sample Output:
Case #1: Alice 18 88888 Billy 24 5888 Bob_Volk 24 5888 Dobby 24 5888 Case #2: Joe_Mike 32 3222 Zoe_Bill 35 2333 Williams 30 -22 Case #3: Anny_Cin 95 999999 Michael 5 300000 Alice 18 88888 Cindy 76 76000 Case #4: None
推荐指数：※
来源：http://pat.zju.edu.cn/contests/pat-a-practise/1055
这道题目本身的逻辑很清楚，就是具体处理的时候考虑其给出的”each contains three positive integers: M (<= 100) - the maximum number of outputs“，感谢：Linest Blog http://linest.github.io/blog/2013/05/23/1/ 关于这个的提醒（本来case 2一直过不了）。这样其实每个年龄的人，多不会超过100，因为多的也不会输出。PAT的有些题目的测试集也不能掉以轻心。
#include <iostream> #include<string.h> #include<stdlib.h> #include<stdio.h> using namespace std; #define N 100001 int n,k; typedef struct people{ char strname[10]; int age; int money; bool visited; }people; people *peop; int query[3]; int compare_money(const void *a ,const void *b) { if(((people *)a)->money>((people *)b)->money ||((people *)a)->age<((people *)b)->age&&((people *)a)->money==((people *)b)->money ||((people *)a)->age==((people *)b)->age&&((people *)a)->money==((people *)b)->money &&strcmp(((people *)a)->strname,((people *)b)->strname)<0) return -1; else return 1; } int main() { int i,j; scanf("%d%d",&n,&k); peop=new people[n]; for(i=0;i<n;i++){ scanf("%s%d%d",peop[i].strname,&peop[i].age,&peop[i].money); } qsort(peop,n,sizeof(people),compare_money); int filter_num=0, filter[N],age_count[201]={0}; for(i=0;i<n;i++){ if(++age_count[peop[i].age]<101)//thank Linest,this line stolen from his blog filter[filter_num++]=i; } for(i=0;i<k;i++){ scanf("%d%d%d",&query[0],&query[1],&query[2]); printf("Case #%d:\n",i+1); int flag=0; int print_num=0; for(j=0;j<filter_num&&print_num<query[0];j++){ int tmp=filter[j]; int age=peop[tmp].age; if(age>=query[1]&&age<=query[2]){ printf("%s %d %d\n",peop[tmp].strname,peop[tmp].age,peop[tmp].money); print_num++; } } if(print_num==0) printf("None\n"); } return 0; }