How Long Does It Take
英文不是很好的我看了好久才知道什么意思
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
1 #include<iostream> 2 #include<vector> 3 #include<queue> 4 using namespace std; 5 #define maxNv 100 6 #define maxtime 100000 7 int flag=1; 8 vector<int> earliest(maxNv,0); 9 queue<int> q; 10 vector<int> rudu(maxNv,0); 11 struct graph{ 12 int Nv,Ne; 13 int G[maxNv][maxNv]; 14 }; 15 using Graph=graph*; 16 struct enode{ 17 int v1,v2; 18 int time; 19 }; 20 using edge=enode*; 21 Graph CreateGraph() 22 { 23 Graph gra=new graph(); 24 cin>>gra->Nv>>gra->Ne; 25 for(int i=0;i<gra->Nv;i++) 26 for(int j=0;j<gra->Nv;j++) 27 gra->G[i][j]=maxtime; 28 return gra; 29 } 30 void Insert(edge e,Graph gra){ 31 gra->G[e->v1][e->v2]=e->time; 32 rudu[e->v2]++; 33 } 34 Graph BuildGraph(){ 35 Graph gra=CreateGraph(); 36 edge e=new enode(); 37 for(int i=0;i<gra->Ne;i++){ 38 cin>>e->v1>>e->v2>>e->time; 39 Insert(e,gra); 40 } 41 return gra; 42 } 43 void timeset(int v1,Graph gra){ 44 int f=0; 45 for(int i=0;i<gra->Nv;i++) 46 if(gra->G[i][v1]!=maxtime&&earliest[i]+gra->G[i][v1]>f) 47 f=earliest[i]+gra->G[i][v1]; 48 earliest[v1]=f; 49 } 50 void solve(Graph gra){ 51 int v1=0,v2=0; 52 for(int i=0;i<gra->Nv;i++) 53 if(rudu[i]==0) { 54 q.push(i); 55 earliest[0]=0; 56 } 57 while(!q.empty()){ 58 v1=q.front(); 59 q.pop(); 60 rudu[v1]--; 61 timeset(v1,gra); 62 for(v2=0;v2<gra->Nv;v2++) 63 if(gra->G[v1][v2]!=maxtime&&rudu[v2]!=-1){ 64 rudu[v2]--; 65 if(rudu[v2]==0) q.push(v2);} 66 } 67 // for(int i=0;i<gra->Nv;i++) 68 // cout<<rudu[i]<<" "; 69 // cout<<endl; 70 for(int i=0;i<gra->Nv;i++) 71 if(rudu[i]!=-1) flag=0; 72 if(flag==0) cout<<"Impossible"; 73 else{ 74 int time=-1; 75 for(int i=0;i<gra->Nv;i++) 76 if(earliest[i]>time) time=earliest[i]; 77 cout<<time; 78 } 79 } 80 int main() 81 { 82 Graph gra=BuildGraph(); 83 solve(gra); 84 return 0; 85 } 86 } } 87 int main() { 88 Graph gra=BuildGraph(); 89 solve(gra); 90 return 0; }
