Spring-outing Decision
Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 676 Accepted Submission(s): 220
is spring ! The sunshine is warm , the flowers is coming out . How
lovely it is! So my classmates and I want to go out for a spring-outing.
But
we all select courses ourselves. We don't have classes at the same
time.Now our monitor has a big trouble in arranging the time of the
spring-outing.
Can you help him?
I will give you our courses information and the time of the spring-outing.You just need to tell me that who can't go with us.
Then CA cases follow.
Each case contains two parts,the students' courses information and the query.
In
the first part ,first there is an integer N（N<200） which means the
number of the student,and then comes the N students’ courses
information.
A student's courses information is in this format:
line1: name K
line2: day1 b1 e1
.....
lineK+1: dayK bK eK
The
first line of a student's courses infomation contains his name(less
than 20 characters and in lowercase) and the number(K,K<1000) of his
courses . Then next K lines describe his courses. Each Line contain
three integers indicate the day of a week( 1 <= day <= 7 means
Monday to Sunday ), the begin time and the end time of the course.
To
make the problem easier,the begin time and the end time will be in the
range from 1 to 11 .(Because in HDU,there is 11 classes one day).
In the query part , first there is an integer Q which means the query number,and then Q lines follow.
A
query contains three integers which means the day ,the begin time and
the end time of the spring-outing.And the time is described as the
courses.
Notice,everyone may have more than one course at the same time for some special reasons.
Please separate two names with a blank.
If all of the students have time to go , just print "None" in a line.
3
linle 3
1 1 2
2 3 4
3 8 10
laili 1
4 1 4
xhd 2
1 2 4
4 5 6
3
1 2 2
4 4 5
5 1 2
laili xhd
None
#include<stdio.h>
#include<string.h>
#include<iostream>
#include<algorithm>
using namespace std;
int Map[][][];
char str[][];
char tstr[][];
bool ans[];
int cnt[];
int main(){
int tt;
scanf("%d",&tt);
while(tt--){
memset(Map,,sizeof(Map));
memset(str,,sizeof(str));
int n;
scanf("%d",&n);
for(int i=;i<=n;i++){
getchar();
int k;
scanf("%s %d",str[i],&k);
for(int j=;j<=k;j++){
int a,b,c;
scanf("%d%d%d",&a,&b,&c);
for(int z=b;z<=c;z++){
Map[i][a][z]=;
}
}
}
int q;
scanf("%d",&q); for(int i=;i<=q;i++){
memset(ans,false,sizeof(ans));
memset(cnt,,sizeof(cnt));
int t,t1,t2;
scanf("%d%d%d",&t,&t1,&t2);
for(int j=;j<=n;j++){
for(int k=t1;k<=t2;k++){
if(Map[j][t][k])
ans[j]=true;
}
}
int Count=;
for(int j=;j<=n;j++){
if(ans[j])
cnt[++Count]=j;
} if(Count==)
printf("None\n");
else{
for(int ii=;ii<Count;ii++){
for(int jj=ii+;jj<=Count;jj++){
if(strcmp(str[cnt[ii]],str[cnt[jj]])>){///此步应该特别注意
int temp1=cnt[ii];
cnt[ii]=cnt[jj];
cnt[jj]=temp1;
}
}
}
for(int qq=;qq<=Count;qq++){
printf("%s%c",str[cnt[qq]],qq==Count?'\n':' ');
}
}
}
}
return ;
}