PAT 1047. Student List for Course

1047. Student List for Course (25)

时间限制

400 ms

内存限制

64000 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 registered course list of each student, you are supposed to output the student name lists of all the courses.

Input Specification:

Each input file contains one test case. For each case, the first line contains 2 numbers: N (<=40000), the total number of students, and K (<=2500), the total number of courses. Then N lines follow, each contains a student's name (3 capital English letters plus a one-digit number), a positive number C (<=20) which is the number of courses that this student has registered, and then followed by C course numbers. For the sake of simplicity, the courses are numbered from 1 to K.

Output Specification:

For each test case, print the student name lists of all the courses in increasing order of the course numbers. For each course, first print in one line the course number and the number of registered students, separated by a space. Then output the students' names in alphabetical order. Each name occupies a line.

Sample Input:

10 5ZOE1 2 4 5ANN0 3 5 2 1BOB5 5 3 4 2 1 5JOE4 1 2JAY9 4 1 2 5 4FRA8 3 4 2 5DON2 2 4 5AMY7 1 5KAT3 3 5 4 2LOR6 4 2 4 1 5

Sample Output:

1 4ANN0BOB5JAY9LOR62 7ANN0BOB5FRA8JAY9JOE4KAT3LOR63 1BOB54 7BOB5DON2FRA8JAY9KAT3LOR6ZOE15 9AMY7ANN0BOB5DON2FRA8JAY9KAT3LOR6ZOE1

刚开始是以string来存储名字，超时

#include <iostream>#include <cstdio>#include <vector>#include <algorithm>using namespace std;int N, C, n, c;string name;int main(){    cin >> N >> C;    // 要声明vv的长度，不然会抛异常    vector < vector <string> > vv(N + 1);    for(int i=0; i<N; i++) {        cin >> name >> n;        for(int j=0; j<n; j++) {            cin >> c;            vv[c].push_back(name);        }    }    for(int i=1; i<=C; i++) {        sort(vv[i].begin(), vv[i].end());        cout << i << " " << vv[i].size() << endl;        for(int j=0; j<vv[i].size(); j++)            cout << vv[i][j] << endl;    }    return 0;}

网上查了一下资料，说是“string的copy时间以及比较时间都要比char数组的相应操作更耗时”，然后就按照Hash的思想吧字符串转为Int整形

#include <iostream>#include <cstdio>#include <vector>#include <algorithm>using namespace std;int N, C, n, c;char name[5];int toInt(char* s) {    return (s[0] - 'A') * 26 * 26 * 26 +           (s[1] - 'A') * 26 * 26 +           (s[2] - 'A') * 26 +           (s[3] - '0');}char* toChars(int i) {    name[0] = (char)(i / 26 / 26 / 26 + 'A');    name[1] = (char)(i / 26 / 26 % 26 + 'A');    name[2] = (char)(i / 26 % 26 + 'A');    name[3] = (char)(i % 26 + '0');    return name;}int main(){    cin >> N >> C;    // 要声明vv的长度，不然会抛异常    vector < vector <int> > vv(N + 1);    for(int i=0; i<N; i++) {        scanf("%s %d", name, &n);        for(int j=0; j<n; j++) {            cin >> c;            vv[c].push_back(toInt(name));        }    }    for(int i=1; i<=C; i++) {        sort(vv[i].begin(), vv[i].end());        printf("%d %d\n", i, vv[i].size());        for(int j=0; j<vv[i].size(); j++)            printf("%s\n", toChars(vv[i][j]));    }    return 0;}

网上也有解法是吧String转为char[]，并把char[]封装到一个结构体中，同时定义sort规则