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_Worth
The 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 50
Sample 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
#include <cstdio> #include <cstring> #include <algorithm> using namespace std; const int maxn=100010; int Age[maxn]={0}; //记录某年龄的人数（下标表示0-200岁） struct Person{ int age, worth; //年龄，财富值 char name[10]; //姓名 }ps[maxn], validps[maxn]; //所有人，能有效输出的人（即在各自年龄段中财富值在100名以内） bool cmp(Person a, Person b){ if(a.worth!=b.worth) return a.worth>b.worth; //按财富值从大到小排序 else if(a.age!=b.age) return a.age<b.age; //按年龄从小到大排序 return strcmp(a.name, b.name)<0; //按姓名字典序从小到大排序 } int main() { int n, k; scanf("%d%d", &n, &k); for(int i=0; i<n; i++){ scanf("%s%d%d", ps[i].name, &ps[i].age, &ps[i].worth); } //步骤(1): 对所有人排序 sort(ps, ps+n, cmp); //步骤(2): m<=100 => //可采用预处理，将同一年龄中财富在前100名以内的人全都存到另一个数组中 //使之后的查询操作均在这个新数组中进行。该预处理将显著降低查询的复杂程度 int validNum=0; //记录存放到valid数组中的人数（validps[]存放能有效输出的人） for(int i=0; i<n; i++){ if(Age[ps[i].age] < 100){ //若年龄为ps[i].age的人数小于100 Age[ps[i].age]++; //年龄ps[i].age的人数+1 validps[validNum++]=ps[i]; //讲ps[i]加入到新数组中 } } //步骤(3): 输出 int m, ageL, ageR; for(int i=0; i<k; i++){ scanf("%d%d%d", &m, &ageL, &ageR); //查询年龄在[ageL, ageR]内的前m个人 printf("Case #%d:\n", i+1); int printNum=0; //已输出的人数（0~m ）中，年龄在当前指定区间的前m个人 //打印 min(validNum, printNum)个 （Person validps[validNum]） for(int j=0; j<validNum && printNum<m; j++){ if(validps[j].age>=ageL && validps[j].age<=ageR){ printf("%s %d %d\n", validps[j].name, validps[j].age, validps[j].worth); printNum++; } } if(printNum==0) printf("None\n"); } return 0; }