08-图8 How Long Does It Take

7-12 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

#include<iostream>
#include<queue>
using namespace std;
const int maxn=105,infinity=10000;
int N,M,minTime[maxn] = {0},InDegree[maxn] = {0},TopNum[maxn] = {0},G[maxn][maxn] = {0};
void ToplogicSort(){
    int v,mark = 1,cnt = 0;
    queue<int>q;
    for(int i=0;i<N;i++){
        if(!InDegree[i]){
            q.push(i);
            cnt++;
        }
    }
    while(!q.empty()){
        v = q.front();
        TopNum[cnt] = v;
        q.pop();
        for(int i=0;i<N;i++){
            if(G[v][i] < infinity){
                InDegree[i]--;
                minTime[i] = max(minTime[v] + G[v][i],minTime[i]);
                if(!InDegree[i]){
                    q.push(i);
                    cnt++;
                }
            }
        }
    }
    int maxT = 0;
    if(cnt != N) cout<<"Impossible"<<endl;
    else{
        for(int i=0;i<N;i++){
            if(minTime[i] > maxT) maxT = minTime[i];
        }
        cout<<maxT<<endl;
    }
}
int main(){
    int a,b,l;
    cin>>N>>M;
    for(int i=0;i<N;i++){
        for(int j=0;j<N;j++){
            G[i][j] = infinity;
        }
    }
    for(int i=0;i<M;i++){
        cin>>a>>b>>l;
        InDegree[b]++;
        G[a][b] = l;
    }
    ToplogicSort();

    return 0;
}