Google codejam 2020 parenting partnering returns solution
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Problem Statement
Cameron and Jamie’s kid is almost 3 years old! However, even though the child is more independent now, scheduling kid activities and domestic necessities is still a challenge for the couple. Cameron and Jamie have a list of N activities to take care of during the day. Each activity happens during a specified interval during the day. They need to assign each activity to one of them, so that neither of them is responsible for two activities that overlap. An activity that ends at time t is not considered to overlap with another activity that starts at time t. For example, suppose that Jamie and Cameron need to cover 3 activities: one running from 18:00 to 20:00, another from 19:00 to 21:00 and another from 22:00 to 23:00. One possibility would be for Jamie to cover the activity running from 19:00 to 21:00, with Cameron covering the other two. Another valid schedule would be for Cameron to cover the activity from 18:00 to 20:00 and Jamie to cover the other two. Notice that the first two activities overlap in the time between 19:00 and 20:00, so it is impossible to assign both of those activities to the same partner. Given the starting and ending times of each activity, find any schedule that does not require the same person to cover overlapping activities, or say that it is impossible.Input
The first line of the input gives the number of test cases, T. T test cases follow. Each test case starts with a line containing a single integer N, the number of activities to assign. Then, N more lines follow. The i-th of these lines (counting starting from 1) contains two integers Si and Ei. The i-th activity starts exactly Si minutes after midnight and ends exactly Ei minutes after midnight.Output
For each test case, output one line containing Case #x: y, where x is the test case number (starting from 1) and y is IMPOSSIBLE if there is no valid schedule according to the above rules, or a string of exactly N characters otherwise. The i-th character in y must be C if the i-th activity is assigned to Cameron in your proposed schedule, and J if it is assigned to Jamie. If there are multiple solutions, you may output any one of them. (See “What if a test case has multiple correct solutions?” in the Competing section of the FAQ. This information about multiple solutions will not be explicitly stated in the remainder of the 2020 contest.)Limits
Time limit: 20 seconds per test set. Memory limit: 1GB. 1 ≤ T ≤ 100. 0 ≤ Si < Ei ≤ 24 × 60.Test set 1 (Visible Verdict)
2 ≤ N ≤ 10.Test set 2 (Visible Verdict)
2 ≤ N ≤ 1000.Sample
| Input | Output |
| 4 3 360 480 420 540 600 660 3 0 1440 1 3 2 4 5 99 150 1 100 100 301 2 5 150 250 2 0 720 720 1440 | Case #1: CJC Case #2: IMPOSSIBLE Case #3: JCCJJ Case #4: CC |
#include <bits/stdc++.h>
using namespace std;
struct Interval {
int start;
int end;
int index;
};
bool compareInterval(Interval i1, Interval i2) {
return (i1.start < i2.start);
}
int main() {
int T, k = 1;
cin >> T;
while (T--) {
int N, S, E, cameron, jamie;
std::string output = "";
bool impossible = false;
cin >> N;
vector <Interval> array(N);
for (int i = 1; i <= N; i++) {
cin >> S >> E;
array[i-1].start = S;
array[i-1].end = E;
array[i-1].index = i-1;
}
// sort the array in terms of increasing start times in the interval
sort(array.begin(), array.end(), compareInterval);
vector<string> schedule(N);
schedule[array[0].index] = "C";
cameron = array[0].end;
schedule[array[1].index] = "J";
jamie = array[1].end;
for (int p = 2; p < N; p++) {
if (cameron <= array[p].start) {
cameron = array[p].end;
schedule[array[p].index] = "C";
}
else if (jamie <= array[p].start) {
jamie = array[p].end;
schedule[array[p].index] = "J";
} else {
impossible = true;
}
}
if (impossible)
output = "IMPOSSIBLE";
else {
for (int x = 0; x < N; ++x)
output += schedule[x];
}
cout << "Case #" << k << ": " << output << endl;
//k is for printing the Case Number
k++;
}
return 0;
}
- Sort the input set of intervals in increasing order of start time. We use a struct Interval which has startTime, endTime and index for saving the schedule map for the given input set of intervals
- Get user input (intervals) and push it into vector of struct Interval. Once all input is read, sort the vector by startTime.
- Keep a schedule vector which will hold the output sequence of persons based on activity. Initialize the first 2 items of this schedule vector to Cameron and Jamie or vice versa as the first 2 activities can be assigned to any of them in any order.
- Save the end time of each of the person’s activities as seen in variables cameron and jamie for itervative comparison. Loop through the remaining set of intervals, compare the start time of each of the intervals with the end times of Cameron and Jamie. If the end time of a particular person is less than the start time of the current activity (interval) being processed, assign it to the respective person, viz. Cameron or Jamie as applicable. If there is an overlap with both of the persons for a given activity, then that means that there is no valid schedule for the given interval set and set the impossible flag to TRUE.
- Once we have processed all the intervals, output the sequence, by iterating through the schedule vector, appending each character in vector to an output string. Once all elements are processed, output the schedule string. If impossible flag is set to true, output “IMPOSSIBLE”
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