7-12 How Long Does It Take (25分) 最短工期

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 (≤), 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 Mlines 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

应该是关键路径的问题，用拓扑排序来解决。
代码：

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int n,m,ans;
int mp[100][100];
int l[100],q[100],t[100];
int main() {
    int a,b,c,head = 0,tail = 0;
    scanf("%d%d",&n,&m);
    memset(mp,-1,sizeof(mp));
    for(int i = 0;i < m;i ++) {
        scanf("%d%d%d",&a,&b,&c);
        mp[a][b] = c;
        l[b] ++;
    }
    for(int i = 0;i < n;i ++) {
        if(!l[i]) {
            q[tail ++] = i;
        }
    }
    while(head < tail) {
        int temp = q[head ++];
        if(t[temp] > ans) ans = t[temp];
        for(int i = 0;i < n;i ++) {
            if(mp[temp][i] != -1) {
                l[i] --;
                if(!l[i]) q[tail ++] = i;
                if(t[i] < t[temp] + mp[temp][i]) {
                    t[i] = t[temp] + mp[temp][i];
                }
            }
        }
    }
    if(tail < n) printf("Impossible");
    else printf("%d",ans);
}