PAT 1039. Course List for Student (25)(倒搜索,注意scanf和string时间消耗大)
来源:互联网 发布:mac os 10.13 开机慢 编辑:程序博客网 时间:2024/06/03 19:03
官网
1039. Course List for Student (25)
时间限制
200 ms
内存限制
65536 kB
代码长度限制
16000 B
判题程序
Standard
作者
CHEN, Yue
Zhejiang University has 40000 students and provides 2500 courses. Now given the student name lists of all the courses, you are supposed to output the registered course list for each student who comes for a query.
Input Specification:
Each input file contains one test case. For each case, the first line contains 2 positive integers: N (<=40000), the number of students who look for their course lists, and K (<=2500), the total number of courses. Then the student name lists are given for the courses (numbered from 1 to K) in the following format: for each course i, first the course index i and the number of registered students Ni (<= 200) are given in a line. Then in the next line, Ni student names are given. A student name consists of 3 capital English letters plus a one-digit number. Finally the last line contains the N names of students who come for a query. All the names and numbers in a line are separated by a space.
Output Specification:
For each test case, print your results in N lines. Each line corresponds to one student, in the following format: first print the student’s name, then the total number of registered courses of that student, and finally the indices of the courses in increasing order. The query results must be printed in the same order as input. All the data in a line must be separated by a space, with no extra space at the end of the line.
Sample Input:
11 5
4 7
BOB5 DON2 FRA8 JAY9 KAT3 LOR6 ZOE1
1 4
ANN0 BOB5 JAY9 LOR6
2 7
ANN0 BOB5 FRA8 JAY9 JOE4 KAT3 LOR6
3 1
BOB5
5 9
AMY7 ANN0 BOB5 DON2 FRA8 JAY9 KAT3 LOR6 ZOE1
ZOE1 ANN0 BOB5 JOE4 JAY9 FRA8 DON2 AMY7 KAT3 LOR6 NON9
Sample Output:
ZOE1 2 4 5
ANN0 3 1 2 5
BOB5 5 1 2 3 4 5
JOE4 1 2
JAY9 4 1 2 4 5
FRA8 3 2 4 5
DON2 2 4 5
AMY7 1 5
KAT3 3 2 4 5
LOR6 4 1 2 4 5
NON9 0
解题思路
- 1.会用
map<int, vector<int> > student
就行了,用它先保存在输出就行了,注意:直接用map<string, vector<int> > student
会超时。
代码
- 1.下面这个居然还会超时,共220毫秒
#include <iostream>#include<map>#include<vector>#include<string>#include<algorithm>using namespace std;int n,k;int w[] = {1,26,26 * 26,26 * 26 * 26};int make(string p){ return (p[3] - '0') * w[0] + (p[2] - 'A') * w[1] + (p[1] - 'A') * w[2] + (p[0] - 'A') * w[3];}map<int,vector<int> > student;int main(){ cin >> n >> k; for (int i = 0; i < k; ++i) { int tem_course,numOfstu; cin >> tem_course >> numOfstu; for (int j = 0; j < numOfstu; ++j) { string strname; cin >> strname; int name =make(strname); student[name].push_back(tem_course); } } for (int i = 0; i < n; ++i) { string str_name; // tem_name.empty() cin >> str_name; int tem_name = make(str_name); vector<int> tem_course = student[tem_name]; cout << str_name << " " << tem_course.size(); if (!tem_course.empty()) { sort(tem_course.begin(),tem_course.end()); for (int j = 0; j < tem_course.size(); ++j) { cout<<" "<< tem_course[j]; } } cout << endl; } return 0;}
- 2.这个不会超时,时间出奇的少。
#include <iostream>#include <cstdio>#include <vector>#include <map>#include <algorithm>using namespace std;map<int,vector<int> > mp;int w[] = {1,26,26 * 26,26 * 26 * 26};int make(char *p){ return (p[3] - '0') * w[0] + (p[2] - 'A') * w[1] + (p[1] - 'A') * w[2] + (p[0] - 'A') * w[3];}int main(){ int n,k,i,j; scanf("%d%d",&n,&k); char name[6]; for(i = 0; i < k; i++) { int c_num,s_num; scanf("%d%d",&c_num,&s_num); for(j = 0; j < s_num; j++) { scanf("%s",name); mp[make(name)].push_back(c_num); } } for(i = 0; i < n; i++) { scanf("%s",name); int ss = make(name); if(mp.find(ss) != mp.end()) { sort(mp[ss].begin(),mp[ss].end()); printf("%s %d",name,mp[ss].size()); for(j = 0; j < mp[ss].size(); j++) printf(" %d",mp[ss][j]); puts(""); } else { printf("%s 0\n",name); } } return 0;}
- PAT 1039. Course List for Student (25)(倒搜索,注意scanf和string时间消耗大)
- 1039. Course List for Student (25)-PAT
- PAT 1039. Course List for Student (25)
- PAT 1039. Course List for Student (25)
- PAT 1039. Course List for Student (25)
- PAT 1039. Course List for Student (25)
- 【PAT】1039. Course List for Student (25)
- PAT 1039. Course List for Student
- PAT 1039. Course List for Student
- PAT-A-1039. Course List for Student
- 【PAT】1039. Course List for Student
- PAT--1039. Course List for Student
- PAT-AL 1039. Course List for Student
- 1047. Student List for Course (25)-PAT
- 【PAT】1047. Student List for Course (25)
- PAT 1047. Student List for Course (25)
- PAT 1007 Course List for Student (25)
- PAT 1047. Student List for Course (25)
- C++中throw异常
- 基于django框架编写的简单信息采集系统
- 通过网络Url下载图片并下载到本地相册
- Android Studio中如何使用SlidingMenu框架
- Strategy_Level1
- PAT 1039. Course List for Student (25)(倒搜索,注意scanf和string时间消耗大)
- 写了一个开机启动,监听通话的demo
- JMeter压测基础篇:linux服务器执行压测三部曲之二
- String拼接符"+"在编译期做了什么?
- Android ListView使用BaseAdapter与ListView的优化
- 【NOIP2016提高A组模拟9.7】总结
- 实现购物网站最近浏览商品的流程
- Office Skills
- 无法控制的并发问题