PAT 1007 Course List for Student (25)

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.

输入描述:

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.

输出描述:

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.

输入例子:

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

输出例子:

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

代码如下

#include <iostream>
#include <cstring>
#include <string>
#include <algorithm>

using namespace std;

int test[26][26][26][10];
int ans[2500*200];

int bs(int key,int l,int r)
{
	int mid ;
	while ( l <= r )
	{
		mid = ( l + r ) >> 1 ;
		if( ans[mid] > key )
			r = mid - 1 ;
		else
			l = mid + 1 ;
	}
	return r ;
}

int main()
{
	int N,K,index,num,i,j,k,m,mm;
	char tname[16];
	cin>>N>>K;

	m=1;
	mm=0;
	for(i=0;i<K;i++)
	{
		cin>>index>>num;
		for(j=0;j<num;j++)
		{
			cin>>tname;
			int &l=test[tname[0]-'A'][tname[1]-'A'][tname[2]-'A'][tname[3]-'0'];
			if(0==l)
				l=m++;
			ans[mm++]=l*2500+index;
 //			cout<<l<<" "<<stu[l].count<<" "<<index<<endl;
		}
	}

	sort(ans,ans+mm);

	for(i=0;i<N;i++)
	{
		cin>>tname;
		int t=test[tname[0]-'A'][tname[1]-'A'][tname[2]-'A'][tname[3]-'0'];
		if(0==t)
		{
			cout<<tname<<" "<<0<<endl;
			continue;
		}
		j = bs(t*2500,0,mm-1) ;
		k = bs((t+1)*2500,j,mm-1);
		cout<<tname<<" "<<k-j;
		for ( j++; j <= k ; j++)
			cout<<" "<<ans[j]-t*2500;
		cout<<endl;
	}
	return 0;
}

这个图片是未去掉代码中注释的这一行

cout<<l<<" "<<stu[l].count<<" "<<index<<endl;
可以看的更加直观

其实AC的代码并不是我最初的思路，按最初的思路写的代码一直没能通过，一时也找不出来原因，以后在看吧

#include <iostream>
#include <cstring>
#include <string>
#include <algorithm>

using namespace std;

typedef struct Node
{
	int count;
	int ans[2500];
}Node;

int test[26][26][26][10];

int main()
{
	int N,K,index,num,i,j,m;
	char tname[16];
	cin>>N>>K;
	Node *stu= new Node[N+1];

	for(i=0;i<N+1;i++)
	{
		stu[i].count=0;
		stu[i].ans[2500]=0;
	}

	m=1;
	for(i=0;i<K;i++)
	{
		cin>>index>>num;
		for(j=0;j<num;j++)
		{
			cin>>tname;
			int &l=test[tname[0]-'A'][tname[1]-'A'][tname[2]-'A'][tname[3]-'0'];
			if(0==l)
				l=m++;
			stu[l].ans[stu[l].count]=index;
			stu[l].count++;
//        cout<<l<<" "<<stu[l].count<<" "<<index<<endl;
		}
	}

	for(i=1;i<N+1;i++)
		sort(stu[i].ans,stu[i].ans+stu[i].count);

	for(i=0;i<N;i++)
	{
		cin>>tname;
		int t=test[tname[0]-'A'][tname[1]-'A'][tname[2]-'A'][tname[3]-'0'];
		if(0==t)
		{
			cout<<tname<<" "<<0<<endl;
			continue;
		}
		cout<<tname<<" "<<stu[t].count;
		for(j=0;j<stu[t].count;j++)
				cout<<" "<<stu[t].ans[j];
		cout<<endl;
	}
	return 0;
}