L

L - Access System

Time Limit:2000MS     Memory Limit:65536KB     64bit IO Format:%lld & %llu

Submit Status Practice ZOJ 3787

Description

For security issues, Marjar University has an access control system for each dormitory building.The system requires the students to use their personal identification cards to open the gate if they want to enter the building.

The gate will then remain unlocked for L seconds. For example L = 15, if a student came to the dormitory at 17:00:00 (in the format of HH:MM:SS) and used his card to open the gate. Any other students who come to the dormitory between [17:00:00, 17:00:15) can enter the building without authentication. If there is another student comes to the dorm at 17:00:15 or later, he must take out his card to unlock the gate again.

There are N students need to enter the dormitory. You are given the time they come to the gate. These lazy students will not use their cards unless necessary. Please find out the students who need to do so.

Input

There are multiple test cases. The first line of input contains an integer T indicating the number of test cases. For each test case:

The first line contains two integers N (1 <= N <= 20000) and L (1 <= L <= 3600). The next N lines, each line is a unique time between [00:00:00, 24:00:00) on the same day.

Output

For each test case, output two lines. The first line is the number of students who need to use the card to open the gate. The second line the the index (1-based) of these students in ascending order, separated by a space.

Sample Input

32 112:30:0012:30:015 1517:00:0017:00:1517:00:0617:01:0017:00:143 512:00:0912:00:0512:00:00

Sample Output

21 231 2 422 3

解题思路：

题意是每个同学刷一次卡能使门禁打开l秒，给出每个同学到门口的时间，统计有多少位同学需要刷卡，并输出其下标

需要把时间化为秒，然后用结构数组存放每位同学的编号和时间，然后对数组用时间排序，定义tt位当前迭代时间，再输出a[0].time后（因为a[0]同学必须刷卡）tt初始化为a[0].time+l,对于找到当前同学后，下一刷卡同学的时间应为当前同学时间+l后出现的第一位同学，对于下标要放入数组中排序输出

心得体会：

此题还算是有点麻烦的，做了很长时间，很多点都需要想到

代码：

#include<bits/stdc++.h>using namespace std;struct stu{int num;int time;};bool comp(const stu &a,const stu & b){return a.time<b.time;}int main(){int n,l,i,t;int h,m,s; char c,cc;stu a[200005];cin>>t;while(t--){vector<int>b;cin>>n>>l;for(i=0;i<n;i++){cin>>h>>c>>m>>cc>>s;a[i].time=3600*h+60*m+s; a[i].num=i+1;}sort(a,a+n,comp);//cout<<a[0].time<<"   "<<a[1].time<<endl; int sum=1;//if((a[n-1].time-a[0].time+1)%l==0) sum=(a[n-1].time-a[0].time+1)/l;//else sum=(a[n-1].time-a[0].time+1)/l+1;//cout<<sum<<endl; int tt=a[0].time+l;b.push_back(a[0].num);for(i=0;i<n;i++){if(a[i].time>=tt) { sum++;int ccc=a[i].num; b.push_back(ccc); tt=a[i].time+l;}}cout<<sum<<endl;sort(b.begin(),b.end());cout<<b[0];for(i=1;i<b.size();i++)cout<<" "<<b[i];cout<<endl;b.clear(); }return 0;}