08-图8 How Long Does It Take (25 分)
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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 (≤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<stdio.h> #include<queue> using namespace std; const int maxn = 110; int map[maxn][maxn],d[maxn]; int inDegree[maxn]; void init(int n); int main() { int n,m; scanf("%d%d",&n,&m); init(n); 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; } } while (!q.empty()) { int cur = q.front(); q.pop(); for (int i = 0; i < n; i++) { if (map[cur][i] != -1) { inDegree[i]--; if (d[i] < d[cur] + map[cur][i]) { d[i] = d[cur] + map[cur][i]; } if (!inDegree[i]) { q.push(i); } } } } int maxCost = -1; bool flag = true; for (int i = 0; i < n; i++) { if (inDegree[i]) { flag = false; break; } if (d[i] > maxCost) { maxCost = d[i]; } } if (flag) { printf("%d",maxCost); } else { printf("Impossible"); } return 0; } void init(int n) { 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; } } }
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