hdu 1497 Simple Library Manageme…
Simple Library Management System
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 756 Accepted Submission(s): 245
Problem Description
After AC all the hardest problems in the world , the ACboy 8006 now has nothing to do . One day he goes to an old library to find a part-time job .It is also a big library which has N books and M users.The user's id is from 1 to M , and the book id is from 1 to N . According to the library rules , every user are only allowed to borrow 9 books .But what surprised him is that there is no computer in the library , and everything is just recorded in paper ! How terrible , I must be crazy after working some weeks , he thinks .So he wants to change the situation .
In the other hand , after 8006's fans know it , they all collect money and buy many computers for the library .
Besides the hardware , the library needs a management program . Though it is just a piece of cake for 8006 , the library turns to you , a excellent programer,for help .
What they need is just a simple library management program . It is just a console program , and have only three commands : Borrow the book , Return the book , Query the user .
1.The Borrow command has two parameters : The user id and the book id
The format is : "B ui bi" (1<=ui<=M , 1<=bi<=N)
The program must first check the book bi wether it's in the library . If it is not , just print "The book is not in the library now" in a line .
If it is , then check the user ui .
If the user has borrowed 9 books already, print "You are not allowed to borrow any more" .
Else let the user ui borrow the book , and print "Borrow success".
2.The Return command only has one parameter : The book id
The format is : "R bi" (1<=bi<=N)
The program must first check the book bi whether it's in the library . If it is , just print "The book is already in the library" . Otherwise , you can return the book , and print "Return success".
3.The Query command has one parameter : The user id
The format is : "Q ui" (1<=ui<=M)
If the number of books which the user ui has borrowed is 0 ,just print "Empty" , otherwise print the books' id which he borrows in increasing order in a line.Seperate two books with a blank.
In the other hand , after 8006's fans know it , they all collect money and buy many computers for the library .
Besides the hardware , the library needs a management program . Though it is just a piece of cake for 8006 , the library turns to you , a excellent programer,for help .
What they need is just a simple library management program . It is just a console program , and have only three commands : Borrow the book , Return the book , Query the user .
1.The Borrow command has two parameters : The user id and the book id
The format is : "B ui bi" (1<=ui<=M , 1<=bi<=N)
The program must first check the book bi wether it's in the library . If it is not , just print "The book is not in the library now" in a line .
If it is , then check the user ui .
If the user has borrowed 9 books already, print "You are not allowed to borrow any more" .
Else let the user ui borrow the book , and print "Borrow success".
2.The Return command only has one parameter : The book id
The format is : "R bi" (1<=bi<=N)
The program must first check the book bi whether it's in the library . If it is , just print "The book is already in the library" . Otherwise , you can return the book , and print "Return success".
3.The Query command has one parameter : The user id
The format is : "Q ui" (1<=ui<=M)
If the number of books which the user ui has borrowed is 0 ,just print "Empty" , otherwise print the books' id which he borrows in increasing order in a line.Seperate two books with a blank.
Input
The input file contains a series of test cases . Please process to the end of file . The first line contains two integers M and N ( 1<= M <= 1000 , 1<=N<=100000),the second line contains a integer C means the number of commands(1<=C<=10000). Then it comes C lines . Each line is a command which is described above.You can assum all the books are in the library at the beginning of each cases.
Output
For each command , print the message which described above .
Please output a blank line after each test.
If you still have some questions , see the Sample .
Please output a blank line after each test.
If you still have some questions , see the Sample .
Sample Input
5 10
9
R 1
B 1 5
B 1 2
Q 1
Q 2
R 5
Q 1
R 2
Q 1
5 10
9
R 1
B 1 5
B 1 2
Q 1
Q 2
R 5
Q 1
R 2
Q 1
Sample Output
The book is already in the library
Borrow success
Borrow success
2 5
Empty
Return success
2
Return success
Empty
The book is already in the library
Borrow success
Borrow success
2 5
Empty
Return success
2
Return success
Empty
#include
#include
#include
struct node
{
int jb[10];
int num;
}ui[1005];
int main()
{
int M,N,n;
int u,b,a[100010];
char ch;
while(scanf("%d%d",&M,&N)!=EOF)
{
memset(a,-1,sizeof(a));
memset(ui,0,sizeof(ui));
scanf("%d",&n);
while(n--)
{
getchar();
scanf("%c",&ch);
if(ch=='B')
{
scanf("%d %d",&u,&b);
if(a[b]!=-1)
{
printf("The book is not in the library now\n");
}
else
{
if(ui[u].num==9)
{
printf("You are not allowed to borrow any more\n");
}
else
{
printf("Borrow success\n");
ui[u].jb[ui[u].num]=b;
ui[u].num++;
a[b]=u;
}
}
}
else if(ch=='R')
{
int j;
scanf("%d",&b);
if(a[b]==-1)
{
printf("The book is already in the library\n");
}
else
{
int temp;
int num;
num=ui[a[b]].num;
for(j=0;j
{
if(ui[a[b]].jb[j]==b)
{
temp=j;
}
}
for(j=temp;j
{
ui[a[b]].jb[j]=ui[a[b]].jb[j+1];
}
ui[a[b]].num--;
a[b]=-1;
printf("Return success\n");
}
}
else if(ch=='Q')
{
scanf("%d",&u);
if(ui[u].num == 0)
{
printf("Empty\n");
}
else
{
int num=ui[u].num;
for(int i=0;i
{
for(int j=num-1;j>i;--j)
{
if(ui[u].jb[i] > ui[u].jb[j])
{
int t = ui[u].jb[i];
ui[u].jb[i]=ui[u].jb[j];
ui[u].jb[j] = t;
}
}
}
for(i=0;i
{
if( i!=0)
printf(" ");
printf("%d",ui[u].jb[i]);
}
printf("\n");
}
}
}
printf("\n");
}
return 0;
}