08-圖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 (≤), 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
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<cstdio> #include<queue> using namespace std; const int maxn = 110; int map[maxn][maxn],d[maxn],indegree[maxn]; int main(){ int n,m; scanf("%d%d",&n,&m); for(int i = 0; i < n; i++){ d[i] = -1; indegree[i] = 0; for(int j = 0; j < n; j++){ map[i][j] = map[j][i] = -1; } } int u,v,w; for(int i = 0; i < m; i++){ scanf("%d%d%d",&u,&v,&w); map[u][v] = w; indegree[v]++; } queue<int> q; for(int i = 0; i < n; i++){ if(!indegree[i]){ q.push(i); d[i] = 0; } } int cur; while(!q.empty()){ cur = q.front(); q.pop(); for(int i = 0; i < n; i++){ if(map[cur][i] != -1){ indegree[i]--; if(d[cur] + map[cur][i] > d[i]){ d[i] = d[cur] + map[cur][i]; } if(!indegree[i]){ q.push(i); } } } } int maxCost = -1; int i; for(i = 0 ; i < n; i++){ if(indegree[i]) break; if(d[i] > maxCost) maxCost = d[i]; } if(i == n) printf("%d",maxCost); else printf("Impossible"); return 0; }