08-图8 How Long Does It Take（25 分）

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:

9 120 1 60 2 40 3 51 4 12 4 13 5 25 4 04 6 94 7 75 7 46 8 27 8 4

Sample Output 1:

18

Sample Input 2:

4 50 1 10 2 22 1 31 3 43 2 5

#include<iostream>#include<fstream>#include<vector>#include<algorithm>#include<string>#include<map>#include<queue>#include<functional>using namespace std;//ifstream inFile("C:\\Users\\DELL\\Desktop\\in.txt", ios::in);const int MaxVertexNum = 100;typedef int Vertex;typedef int WeightType;typedef char DatatType;//边的定义typedef struct ENode *PtrToENode;struct ENode {Vertex V1, V2;WeightType Weight;};typedef PtrToENode Edge;//邻接点的定义typedef struct AdjVNode *PtrToAdjVNode;struct AdjVNode {WeightType Weight;int earlisttime;//最早完成时间int lasttime;//最后完成时间Vertex AdjV;PtrToAdjVNode Next;};//顶点表头结点的定义typedef struct VNode {int earlisttime;//最早完成时间PtrToAdjVNode FirstEdge;DatatType Data;}AdjList[MaxVertexNum];//图结点的定义typedef struct GNode* PtrToGNode;struct GNode {int Nv;int Ne;AdjList G;};typedef  PtrToGNode LGraph;LGraph CreateGraph(int VertexNum){LGraph Graph = new GNode;Graph->Nv = VertexNum;Graph->Ne = 0;for (int i = 0; i < Graph->Nv; ++i) {Graph->G[i].FirstEdge = nullptr;Graph->G[i].earlisttime = 0;}return Graph;}//插入有向边<V1,V2>void InsertEdge(LGraph Graph, Edge E){PtrToAdjVNode p = new AdjVNode;p->AdjV = E->V2;p->Weight = E->Weight;p->earlisttime = 0;p->Next = Graph->G[E->V1].FirstEdge;Graph->G[E->V1].FirstEdge = p;}LGraph BulidGraph(){int Nv;cin >> Nv;LGraph Graph = CreateGraph(Nv);cin >> Graph->Ne;if (Graph->Ne) {Edge E = new ENode;for (int i = 0; i < Graph->Ne; ++i) {cin >> E->V1 >> E->V2 >> E->Weight;InsertEdge(Graph, E);}}return Graph;}void Print(LGraph Graph){if (Graph->Nv) {for (int i = 0; i < Graph->Nv; ++i) {PtrToAdjVNode p = Graph->G[i].FirstEdge;while (p) {cout << p->AdjV << " ";p = p->Next;}cout << endl;}}}bool TopSort(LGraph Graph, Vertex TopOrder[],int& ret,int m[]){ret = 0;//对Graph进行拓扑排序，TopOrder[]顺序存储排序后的顶点下标int Indegree[MaxVertexNum], cnt;queue<int> Q;//初始化Indegree[]for (int V = 0; V < Graph->Nv; ++V) {Indegree[V] = 0;}//遍历图得到Indegree[]for (int V = 0; V < Graph->Nv; ++V) {for (PtrToAdjVNode W = Graph->G[V].FirstEdge; W; W=W->Next ) {Indegree[W->AdjV]++;//对有向边<V，W->AdjV>累计终点的入度}}//将所有入度为0的点入队for (int V = 0; V < Graph->Nv; ++V) {if (Indegree[V] == 0) {Q.push(V);}}//下面进行拓扑排序cnt = 0;while (Q.size()) {int V = Q.front();//弹出一个入度为0的点Q.pop();TopOrder[cnt++] = V;//将之存为结果序列的下一个元素//对V的每个邻接点W->AdjVfor (PtrToAdjVNode W = Graph->G[V].FirstEdge; W; W = W->Next) {if (--Indegree[W->AdjV] == 0) {//若删除V使得W->AdjV入度为零Q.push(W->AdjV );}if (Graph->G[V].earlisttime + W->Weight >W->earlisttime) {W->earlisttime = Graph->G[V].earlisttime  + W->Weight;Graph->G[W->AdjV ].earlisttime = W->earlisttime;m[W->AdjV] = ret;if (ret < W->earlisttime) {ret = W->earlisttime;}}}}if (cnt != Graph->Nv) {return false;//说明图中有回路，返回不成功的标志}else {return true;}}int main(){LGraph Graph = BulidGraph();//Print(Graph);Vertex TopOrder[MaxVertexNum] ;Vertex m[MaxVertexNum];int ret = 0;if (TopSort(Graph, TopOrder,ret,m)) {cout << ret << endl;}else {cout << "Impossible" << endl;}//delete Graph;//inFile.close();system("pause");return 0;}