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<stdio.h> #include<queue> #include<vector> 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); vector<int>adj[n]; for(i=0;i<m;i++){ scanf("%d %d %d",&c1,&c2,&time); indegree[c2]++; adj[c1].push_back(c2); maptime[c1][c2]=time; } queue<int>q; for(i=0;i<n;i++){ if(indegree[i]==0){ q.push(i); } } int cou=0; while(!q.empty()){ cou++; int head=q.front(); q.pop(); for(i=0;i<adj[head].size();i++){ int next=adj[head][i]; if(timesum[next]<timesum[head]+maptime[head][next]){//每个节点一定是最大值 timesum[next]=timesum[head]+maptime[head][next]; } if(--indegree[next]==0){ q.push(next); } } } if(cou!=n){ printf("Impossible\n"); } else{ int max=-1; for(i=0;i<n;i++){ if(timesum[i]>max){ max=timesum[i]; } } printf("%d",max); } }