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#include <bits/stdc++.h>
using namespace std;
class job
{
public:
    int st, ed, cost;
};
int getTime(string s)
{
    int hr = (s[0] – ‘0’) * 10 + (s[1] – ‘0’);
    int min = (s[2] – ‘0’) * 10 + (s[3] – ‘0’);
 
    return hr * 60 + min;
}
bool compare(job A, job B)
{
    return A.ed < B.ed;
}
int searchJob(job arr[], int st, int ed, int key)
{
    int ans = –1;
    while (st <= ed)
    {
        int mid = (st + ed) / 2;
        if (arr[mid].ed <= key)
        {
            ans = mid;
            st = mid + 1;
        }
        else
        {
            ed = mid – 1;
        }
    }
    return ans;
}
pair<int, int> solve(job arr[], int n)
{
    int dp[n] = {0};
    int numOfJobs[n] = {0};
 
    dp[0] = arr[0].cost;
    numOfJobs[0] = 1;
 
    for (int i = 1; i < n; i++)
    {
        int cur = arr[i].cost;
        int num = 1;
        int idx = searchJob(arr, 0, i – 1, arr[i].st);
 
        if (idx != cur)
        {
            cur += dp[idx];
            num += numOfJobs[idx];
        }
        if (cur > dp[i – 1])
        {
            dp[i] = cur;
            numOfJobs[i] = num;
        }
        else
        {
            dp[i] = dp[i – 1];
            numOfJobs[i] = numOfJobs[i – 1];
        }
    }
    return {numOfJobs[n – 1], dp[n – 1]};
}
int main()
{
    int n;
    cin >> n;
 
    job arr[n];
    int cost;
    string st, ed;
    int total = 0;
 
    for (int i = 0; i < n; i++)
    {
        cin >> st >> ed >> cost;
        arr[i].st = getTime(st);
        arr[i].ed = getTime(ed);
        arr[i].cost = cost;
        total += cost;
    }
    sort(arr, arr + n, compare);
 
    pair<int, int> res = solve(arr, n);
 
    cout << n – res.first << endl;
    cout << total – res.second << endl;
 
    return 0;
}


What I have tried:

C++
#include <stdio..h>
int getTime(string s)
{
    int hr = (s[0] – ‘0’) * 10 + (s[1] – ‘0’);
    int min = (s[2] – ‘0’) * 10 + (s[3] – ‘0’);
 
    return hr * 60 + min;
}
int searchJob(job arr[], int st, int ed, int key)
{
    int ans = –1;
    while (st <= ed)
    {
        int mid = (st + ed) / 2;
        if (arr[mid].ed <= key)
        {
            ans = mid;
            st = mid + 1;
        }
        else
        {
            ed = mid – 1;
        }
    }
    return ans;
}
pair<int, int=""> solve(job arr[], int n)
{
    int dp[n] = {0};
    int numOfJobs[n] = {0};
 
    dp[0] = arr[0].cost;
    numOfJobs[0] = 1;
 
    for (int i = 1; i < n; i++)
    {
        int cur = arr[i].cost;
        int num = 1;
        int idx = searchJob(arr, 0, i – 1, arr[i].st);
 
        if (idx != cur)
        {
            cur += dp[idx];
            num += numOfJobs[idx];
        }
        if (cur > dp[i – 1])
        {
            dp[i] = cur;
            numOfJobs[i] = num;
        }
        else
        {
            dp[i] = dp[i – 1];
            numOfJobs[i] = numOfJobs[i – 1];
        }
    }
    return {numOfJobs[n – 1], dp[n – 1]};
}
int main()
{
    int n;
    cin >> n;
 
    job arr[n];
    int cost;
    string st, ed;
    int total = 0;
 
    for (int i = 0; i < n; i++)
    {
        cin >> st >> ed >> cost;
        arr[i].st = getTime(st);
        arr[i].ed = getTime(ed);
        arr[i].cost = cost;
        total += cost;
    }
    sort(arr, arr + n, compare);
 
    pair<int, int=""> res = solve(arr, n);
 
    cout << n – res.first << endl;
    cout << total – res.second << endl;
 
    return 0;
}
Posted
Updated 26-Feb-23 8:51am
v2
Comments
0x01AA 26-Feb-23 13:35pm    
And, where is the problem exactly?

This is not a code translation service: even if it was, converting C++ to C will not give you "good" C code.
C is an old language from which C++ was evolved: it doesn't contain much of what C++ does at all - classes, vectors, tuples for example are all C++ only.
And while it is possible to recode cin and cout to their more primitive C equivalents, replacing the other C++ specific features would produce horrible complicated code to do a simple task.

Instead, throw that away, and write your own C code to do the job - you'll learn things and produce better code.
 
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Comments
Nelek 26-Feb-23 14:20pm    
:thumbsup:
There are some wrong characters here that don't make sense as code.
Strings are encoded as a char array.
C
//#include <stdio..h>
#include <stdio.h>

//int getTime(string s)
int getTime(char* s)
{
	if (strlen(s) < 4)
		return 0;
	// int hr = (s[0] – ‘0’) * 10 + (s[1] – ‘0’);
	int hr = (s[0] - '0') * 10 + (s[1] - '0');
	// int min = (s[2] – ‘0’) * 10 + (s[3] – ‘0’);
	int min = (s[2] - '0') * 10 + (s[3] - '0');

	return hr * 60 + min;
}
 
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