1055. The World's Richest (25)
2013-09-29 18:48
483 查看
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
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:
Sample Output:
[/code]
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
思路:因为只有200个年龄,输入的行数最长有100000,平均每个年龄有500.加上M最大只有100,所以有些数据需要剔除。
注释部分是超时的代码(test 2不能通过),主要原因是每输入一个M min max,就要把一些vector排序,有重复。
/* #include <iostream> #include"stdio.h" #include"stdlib.h" #include"string.h" #include"algorithm" #include"vector" using namespace std; struct person{ char name[9]; int age; long worth; }; bool comparePersonByWorth(struct person p1, struct person p2){ if(p1.worth>p2.worth){ return true; } return false; } bool compare(struct person p1,struct person p2){ if(p1.worth>p2.worth) return true; if(p1.worth==p2.worth){ if(p1.age<p2.age) return true; if(p1.age==p2.age){ if(strcmp(p1.name,p2.name)<0){ return true; } } } return false; } int main() { vector<struct person> v[201]; long N; int K; int M,minn,maxn; scanf("%ld%d",&N,&K); struct person p; for(long i=0;i<N;i++){ scanf("%s%d%ld",p.name,&p.age,&p.worth); v[p.age].push_back(p); } for(long i=1;i<=200;i++){ if(v[i].size()>100) sort(v[i].begin(),v[i].begin(),comparePersonByWorth); } int v_size,length,s; for(int i=1;i<=K;i++){ scanf("%d%d%d",&M,&minn,&maxn); vector<struct person> v2 = v[minn]; for(int j=minn+1;j<=maxn;j++){ v_size = v[j].size(); length = (100>v_size)?v_size:100; v2.insert(v2.end(),v[j].begin(),v[j].begin()+length); } printf("Case #%d:\n",i); if(!v2.empty()){ sort(v2.begin(),v2.end(),compare); v_size = v2.size(); s = (M>v_size)?v_size:M; for(int k=0;k<s;k++){ printf("%s %d %ld\n",v2[k].name,v2[k].age,v2[k].worth); } } else { printf("None\n"); } } return 0; } */ #include <iostream> #include"stdio.h" #include"stdlib.h" #include"string.h" #include"algorithm" #include"vector" using namespace std; struct person{ char name[9]; int age; long worth; }; bool comparePersonByWorth(struct person p1, struct person p2){ if(p1.worth>p2.worth){ return true; } return false; } bool compare(struct person p1,struct person p2){ if(p1.worth>p2.worth) return true; if(p1.worth==p2.worth){ if(p1.age<p2.age) return true; if(p1.age==p2.age){ if(strcmp(p1.name,p2.name)<0){ return true; } } } return false; } int main() { vector<struct person> v[201]; long N; int K; int M,minn,maxn; scanf("%ld%d",&N,&K); struct person p; for(long i=0;i<N;i++){ scanf("%s%d%ld",p.name,&p.age,&p.worth); v[p.age].push_back(p); } vector<struct person> v2; int v_size,length,s; for(long i=1;i<=200;i++){ /*如果相同年龄的富豪超过100个,则按照财富值筛选出前100*/ if(v[i].size()>100){ sort(v[i].begin(),v[i].end(),comparePersonByWorth); } v_size = v[i].size(); length = (100>v_size)?v_size:100; v2.insert(v2.end(),v[i].begin(),v[i].begin()+length); } sort(v2.begin(),v2.end(),compare); for(int i=1;i<=K;i++){ scanf("%d%d%d",&M,&minn,&maxn); bool flag = false; printf("Case #%d:\n",i); int cnt = 0;//记录已经输出的富豪数目 for(int k=0;cnt<M&&k<v2.size();k++){ if(v2[k].age>=minn&&v2[k].age<=maxn){ printf("%s %d %ld\n",v2[k].name,v2[k].age,v2[k].worth); flag = true; cnt++; } } if(flag==false) { printf("None\n"); } } return 0; }
[/code]
相关文章推荐
- Xcode开发技巧之code snippets(代码片段)
- 教你如何做简单的数据分析--转载微博
- Java注释@interface的用法
- 一步步学算法(算法分析)---6(贪心算法)
- Windows核心编程-----线程池二
- 如何在开发项目里进行自我激励!
- [Linux项目实践] 物联网单板测试:OLED Dis_Menu
- Dreamweaver格式化HTML代码
- Android Tombstone/Crash的log分析和定位
- CF202 Div 1 题解
- Cocos2d-x VS2012 HelloCpp调试
- Windows核心编程-----线程池
- 我爱说英语之学美语发音
- 只能输入IP打开主页,而localhost不行解决办法
- [Linux项目实践] 物联网单板测试之任务三:OLED菜单控制LED
- 合并DataSet 建立treeview多级菜单
- log打印
- PHP执行shell脚本
- 不支持C++11 decltype的噩耗
- 阿里巴巴面试2014--武汉