PAT甲級1039

1039. Course List for Student (25)

時間限制

200 ms

記憶體限制

65536 kB

代碼長度限制

16000 B

判題程式

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

/*
#include<string>
#include<cstdio>
#include<set>
#include<map>
#include<iostream>
using namespace std;

int main()
{
  int N, K;
  //cin >> N >> K;
  scanf("%d %d", &N, &K);
  map<string, set<int> > m;
  int coursei, Ni; char s[5];
  for (int i = 0; i < K; i++)
  {
    //cin >> coursei >> Ni;
    scanf("%d %d", &coursei, &Ni);
    for (int j = 0; j < Ni; j++)
    {
      //cin >> s;
      scanf("%s", s);
      m[s].insert(coursei);
    }
  }
  for (int i = 0; i < N; i++)
  {
    //cin >> s;
    scanf("%s", s);
    //cout << s << " " << m[s].size();
    printf("%s %d", s, m[s].size());
    if (m[s].size())
    {
      for (set<int>::iterator it = m[s].begin(); it != m[s].end(); it++)
      {
        //cout << " " << *(it);
        printf(" %d", *(it));
      }
    }
    //cout << endl;
    printf("\n");
  }
  return 0;
}
用map最後一個測試點過不了，會逾時.順便提一下char數組指派給string時會被自動包裝*/
#include<cstdio>
#include<cstring>
#include<vector>
#include<algorithm>
using namespace std;

const int N = 40010;//總人數
const int M = 26 * 26 * 26*10+ 1;//由姓名散列成的數字上界
vector<int> selectCourse[M];//每個學生選擇的課程編号

int getID(char name[])//hash函數，将字元串name轉換成數字
{
  int id = 0;
  for (int i = 0; i < 3; i++)
  {
    id = id * 26 + (name[i] - 'A');
  }
  id = id * 10 + (name[3] - '0');
  return id;

}
int main()
{
  char name[5];
  int n, k;
  scanf("%d%d", &n, &k);//人數及課程數
  for (int i = 0; i < k; i++)//對每門課程
  {
    int course, x;
    scanf("%d%d", &course, &x);//輸入課程編号與選課人數
    for (int j = 0; j < x; j++)
    {
      scanf("%s", name);
      int id = getID(name);//将名字散列為一個整數作為編号
      selectCourse[id].push_back(course);
    }
  }
  for (int i = 0; i < n; i++)//n個查詢
  {
    scanf("%s", name);
    int id = getID(name);//獲得學生編号
    sort(selectCourse[id].begin(), selectCourse[id].end());//從小到大排序
    printf("%s %d", name, selectCourse[id].size());//姓名、選課數
    for (int j = 0; j < selectCourse[id].size(); j++)
    {
      printf(" %d", selectCourse[id][j]);//選課編号
    }
    printf("\n");
  }
  return 0;
}