Greedy algorithm solves the problem of venue arrangement

Greedy algorithm solves the problem of venue arrangement.

"Problem description"

Suppose you want to arrange a number of events in enough venues and want to use as few venues as possible. (This problem is actually a well-known graph coloring issue.) If each activity is a vertex of the graph, the incompatible activities are connected by edges. Causes the adjacent vertices to have a minimum number of colors, corresponding to the minimum number of venues to find. ）

"Data Entry"

The input data is given by the file Input.txt, and the first line is a positive integer k, indicating that there are k pending activities.

Next there are k rows of data, each with two positive integers representing the start and end times of the K-pending activity.

"End Output"

Output minimum number of venues.

Input.txt output.txt

5 3

1 23

12 28

25 35

27 80

36 50

Idea: Traverse all the activity time, if the time can be superimposed to the previous activity of the venue, then superimposed to the existing activities, if not superimposed, the new venue.

Code:

#include "iostream" #include "string" using namespace std;
	struct TT {int st;
int Ed;


};
	int main () {int n=0;
	FILE *FP;
	int num=1;
	if ((Fp=fopen ("Input.txt", "R")) ==null) {printf ("Can ' t find the input.txt\n");

	} fscanf (FP, "%d", &n);
	int *s_t;
    int *e_t;
	s_t = (int *) malloc (sizeof (int) *n+1);
    e_t = (int *) malloc (sizeof (int) *n+1);
	TT *t_c;

	T_c = (TT *) malloc (sizeof (TT) *n+1);
	int i=0;
	for (i=0;i<n;i++) {fscanf (FP, "%d%d", &s_t[i],&e_t[i]);
	} for (i=0;i<n;i++) {printf ("%3d", S_t[i]);
	} cout<<endl;
	for (i=0;i<n;i++) {printf ("%3d", E_t[i]);




	} cout<<endl;
	int star =1000,end=0;
	int S_key,e_key;
			for (i=0;i<n;i++) {if (S_t[i]<=star) {s_key = i;
		Star = S_t[i];
			} if (e_t[i]>=end) {e_key = i;
		end = E_t[i];
	}} int count=0;
	T_c[0].ed = E_t[s_key];

	Count + +;
		for (i=0;i<n;i++) {int KK = s_t[i];
		if (I==s_key) {continue; } for (int j=0;j<count;j++{if (T_C[J].ED&LT;KK) {T_c[j].ed = E_t[i];
			Break
				} else {T_c[count].ed = e_t[i];
				count++;
			Break







    }}} printf ("%3d\n", count);
return 0; }