08-图8 How Long Does It Take (25分)

Given the relations of all the activities of a project, you are supposed to find the earliest completion time of the project.

Input Specification:

Each input file contains one test case. Each case starts with a line containing two positive integers $N$ ($\le 100$), the number of activity check points (hence it is assumed that the check points are numbered from 0 to $N-1$), and $M$, the number of activities. Then $M$ lines follow, each gives the description of an activity. For the i-th activity, three non-negative numbers are given: S[i], E[i], and L[i], where S[i] is the index of the starting check point, E[i] of the ending check point, and L[i] the lasting time of the activity. The numbers in a line are separated by a space.

Output Specification:

For each test case, if the scheduling is possible, print in a line its earliest completion time; or simply output "Impossible".

Sample Input 1:

9 12
0 1 6
0 2 4
0 3 5
1 4 1
2 4 1
3 5 2
5 4 0
4 6 9
4 7 7
5 7 4
6 8 2
7 8 4

Sample Output 1:

18

Sample Input 2:

4 5
0 1 1
0 2 2
2 1 3
1 3 4
3 2 5

Sample Output 2:

Impossible

拓扑排序，好好背模板吧

#include
#include
#include
using namespace std;
int maptime[100][100]={0};//c1-c2的时间 
int indegree[100]={0};
int timesum[100]={0};
int main(){
	int n,m,i,j,c1,c2,time;
	scanf("%d %d",&n,&m);
	vectoradj[n];
	for(i=0;iq;
	for(i=0;imax){
				max=timesum[i];
			}
		}
		printf("%d",max);
	}
}