【算法笔记6.1小节 - vector 】问题 A: Course List for Student (25)

问题 A: Course List for Student (25)

时间限制: 1 Sec 内存限制: 32 MB
提交: 695 解决: 198
[提交][状态][讨论版][命题人:外部导入]

题目描述

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.

输入

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.

输出

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.

样例输入

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

样例输出

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

建立名字和整数之间的映射。

#include<stdio.h>
#include<vector>
#include<string.h>
#include<algorithm>
using namespace std;
//将名字映射为26进制
//前三个字符转换为26进制*10 + 最后一位数字
int change(char a[])
{
    int ans = (a[0]-'A')*26*26*26+(a[1]-'A')*26*26+(a[2]-'A')*26+(a[3]-'0');
    return ans;
}
vector<int>ans[1000000];
int main()
{
    int n, k;
    scanf("%d %d",&n, &k);
    char a[5];
    int temp;
    for(int j=0; j<k; j++)
    {
        int cnum, sum;
        scanf("%d %d", &cnum, &sum);
        for(int i=0; i<sum; i++)
        {
            scanf("%s", a);
            temp = change(a);
            ans[temp].push_back(cnum);
        }
    }
 
 
    for(int i=0; i<n; i++)
    {
        scanf("%s",a);
        temp = change(a);
        sort(ans[temp].begin(),ans[temp].end());
        printf("%s %d",a, ans[temp].size());
        for(int j=0; j< ans[temp].size(); j++)
            printf(" %d",ans[temp][j]);
          printf("\n");
    }
    return 0;
}
 
/**************************************************************
    Problem: 6114
    User: 151210132
    Language: C++
    Result: 正确
    Time:144 ms
    Memory:30284 kb
****************************************************************/