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 NN (le 100≤100), the number of activity check points (hence it is assumed that the check points are numbered from 0 to N-1N−1), and MM, the number of activities. Then MM 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<malloc.h> 3 #include<queue> 4 using namespace std; 5 6 #define INF (100000) 7 #define MAX 100 8 //如果采用邻接链表的形式构造图，在每一个顶点中接下存储的边是其相邻的边 9 typedef struct _Edge{ 10 int Adjv; 11 int Weight; 12 13 14 _Edge *Next; 15 }Edge,nedge; 16 17 typedef struct _VNode 18 { 19 int NodeName; 20 21 nedge *first_edge; 22 }VNode; 23 24 typedef struct _Graph{ 25 int Vernum; 26 int Edgnum; 27 VNode G[MAX]; 28 29 }MGraph; 30 31 MGraph *Graph; 32 33 void link_last(nedge *list, nedge *node) 34 { nedge *p=list; 35 while(p->Next!=NULL) 36 { 37 p=p->Next; 38 } 39 p->Next=node; 40 node->Next=NULL; 41 } 42 43 MGraph *BuildMyGraph(int N,int M) 44 { MGraph *pG; 45 pG=(MGraph*)malloc(sizeof( _Graph));//pG图必须要申请空间 46 47 pG->Vernum=N;pG->Edgnum=M; 48 49 //将点初始化 50 for(int i=0;i<N;i++) 51 { 52 pG->G[i].NodeName=i; 53 54 pG->G[i].first_edge=NULL; 55 } 56 //输入数据 57 int c1,c2,w1,i,j; 58 Edge *edge; 59 for(i=0;i<pG->Edgnum;i++) 60 { 61 cin>>c1>>c2>>w1; 62 edge=(Edge*)malloc(sizeof(_Edge)); 63 edge->Adjv=c2;edge->Weight=w1;edge->Next=NULL; 64 for(j=0;j<pG->Vernum;j++) 65 { 66 if(c1==pG->G[j].NodeName) 67 { 68 if(pG->G[j].first_edge==NULL) 69 { 70 pG->G[j].first_edge=edge; 71 } 72 else 73 { 74 link_last(pG->G[j].first_edge,edge); 75 } 76 } 77 } 78 } 79 return pG; 80 } 81 int TopSort(int Total) 82 {int Indegree[MAX],V;int TopOrder[MAX],i;int S[MAX]; 83 queue<int>Q; Edge *E; 84 85 for( i=0;i<Graph->Vernum;i++) 86 Indegree[i]=0; 87 //计算其入度，每个顶点的入度 88 for(i=0;i<Graph->Vernum;i++) 89 { 90 E=Graph->G[i].first_edge; 91 while(E!=NULL) 92 { 93 Indegree[E->Adjv]++; 94 E=E->Next; 95 96 } 97 } 98 //找到入度为零的点 99 for(i=0;i<Graph->Vernum;i++) 100 { 101 if(Indegree[i]==0) 102 { Q.push(i); 103 S[i]=0; 104 } 105 } 106 int cnt=0; int temp=0; 107 while(!Q.empty()) 108 { 109 V=Q.front();Q.pop(); 110 TopOrder[cnt++]=V; 111 for(E=Graph->G[V].first_edge;E;E=E->Next) 112 { 113 temp=E->Weight; 114 //这里只有一句话，只能计算工程完成要多久，如果深入，编写出其 115 //是否有机动组 116 S[E->Adjv]=S[V]+temp> S[E->Adjv]?S[V]+temp:S[E->Adjv]; 117 if(--Indegree[E->Adjv]==0) 118 Q.push(E->Adjv); 119 } 120 } 121 if(cnt!=Graph->Vernum) 122 Total=-1; 123 else 124 Total=S[(cnt-1)]; 125 return Total; 126 } 127 int main() 128 { 129 int N,M; 130 cin>>N>>M; 131 Graph=BuildMyGraph(N,M); 132 int Total=0; 133 Total =TopSort(Total); 134 if(Total!=-1) 135 { 136 cout<<Total<<endl; 137 } 138 else 139 { 140 cout<<"Impossible"<<endl; 141 } 142 return 0; 143 }