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 10 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 }