HDU-1707-Spring-outing Decision

妖狐艹你老母 2022-05-28 01:10 76阅读 0赞

# Spring-outing Decision #

## Problem Description ##

Now 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.

## Input ##

The first line contains an integer CA which indicates the number of test cases. 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.

## Output ##

For each query , just print the names of the students who can't go out for a spring-outing in a line in lexicographic order. Please separate two names with a blank. If all of the students have time to go , just print "None" in a line.

## Sample Input ##

1 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

## Sample Output ##

linle xhd laili xhd None

#include<bits/stdc++.h>
    using namespace std;
    struct classes
    {
        char a[50];
        int b;
        int c[1005],d[1005],e[1005];
        int f;
    }q[205];
    int cmp(classes x,classes y)
    {
        if(strcmp(x.a,y.a)>0)
        return 0;
        else
        return 1;
    }
    int main()
    {
        int n,a,t,flag,m,c,d,e;
        scanf("%d",&t);
        while(t--)
        {
            scanf("%d",&n);
            for(int i=0;i<n;i++)
            {
                scanf("%s",q[i].a);
                scanf("%d",&q[i].b);
                for(int j=0;j<q[i].b;j++)
                {
                    scanf("%d %d %d",&q[i].c[j],&q[i].d[j],&q[i].e[j]);
                }
            }
            scanf("%d",&m);
            for(int i=0;i<m;i++)
            {
                for(int j=0;j<n;j++)
                {
                    q[j].f=0;
                }
                flag=0;
                scanf("%d%d%d",&c,&d,&e);
                for(int j=0;j<n;j++)
                {
                    for(int k=0;k<q[j].b;k++)
                    {
                        if(c==q[j].c[k])
                        {
                            if((d>=q[j].d[k]&&d<=q[j].e[k])||(e>=q[j].d[k]&&e<=q[j].e[k])||(d<=q[j].d[k]&&e>=q[j].e[k])||(e==q[j].d[k])||(d==q[j].e[k]))
                            {
                                flag=1;
                                q[j].f=1;
                                break;
                            }
                        }
                    }
                }
                if(flag==0)
                {
                    printf("None\n");
                }
                else
                {
                    sort(q,q+n,cmp);
                    for(int l=0;l<n;l++)
                    {
                        if(q[l].f==1)
                        {
                            q[l].f=0;
                            printf("%s",q[l].a);
                            break;
                        }
                    }
                    for(int l=0;l<n;l++)
                    {
                        if(q[l].f==1)
                        {
                            printf(" %s",q[l].a);
                        }
                    }
                    printf("\n");
                }
            }
        }
        return 0;
    }