PTA 08-图8 How Long Does It Take（25 分）

08-图8 How Long Does It Take（25 分）

题目地址：08-图8 How Long Does It Take（25 分）

题目描述：

Given the relations of all the activities of a project, you are supposed to find the earliest completion time of the project.

• 输入格式：
  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.

• 输出格式：
  For each test case, if the scheduling is possible, print in a line its earliest completion time; or simply output “Impossible”.

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Tips:
这题考查的是拓扑排序。会写拓扑排序应该没问题，但是有两个易错点需要注意：
1. 最终完工时间未必是最后一个结点完成的时间，因为可能有多个“结束结点”，最终的完成时间取决于这些结点中时间最长的一个。
2. 我在下面程序中做了标记（########），在我标记的for循环内，千万不要拿来更新EarlyVertexNum，我一开始就是这么写的，结果有两个测试样例通不过，我来我想了一下，如果对题目给的输入样例以来说，倒着输结点之间的权重，更新就会出问题。

---

程序：

#include <iostream>
#include <queue>
#include <cstdlib>
#include <cstdio>
using namespace std;

#define MaxVertexNum 105
typedef int WeightType;
typedef int Vertex;
int EarlyStart[MaxVertexNum] = {0};

/* 边定义 */
typedef struct ENode* Edge;
struct ENode
{
    Vertex V1, V2;
    WeightType Weight;
};

/* 邻接点定义 */
typedef struct AdjNode* PtrToAdjNode;
struct AdjNode
{
    Vertex AdjV;    // 邻接点下标
    WeightType Weight;
    PtrToAdjNode next;  
};

/* 顶点表头结点定义*/
typedef struct VNode
{
    PtrToAdjNode FirstEdge;
}AdjList[MaxVertexNum];

typedef struct GNode* LGraph;
struct GNode
{
    int Nv;     // 顶点数
    int Ne;     // 边数
    AdjList G;  
};

void InsertEdge(Edge E, LGraph Graph)
{   /* 头插法 插入边<V1, V2> */
    PtrToAdjNode NewNode = (PtrToAdjNode)malloc(sizeof(AdjNode));
    NewNode->AdjV = E->V2;
    NewNode->Weight = E->Weight;
    NewNode->next = Graph->G[E->V1].FirstEdge;
    Graph->G[E->V1].FirstEdge = NewNode;
}

LGraph InitialGraph(int Nv)
{   /* 初始化图 */
    LGraph Graph = (LGraph)malloc(sizeof(GNode));
    Graph->Nv = Nv;
    Graph->Ne = 0;
    for (int i = 0; i < Nv; i++)
        Graph->G[i].FirstEdge = NULL;
    return Graph;
}

LGraph BuildGraph(LGraph Graph)
{   /* 建立边联系 */
    cin >> Graph->Ne;
    for (int i = 0; i < Graph->Ne; i++)
    {
        Edge E = (Edge)malloc(sizeof(ENode));
        cin >> E->V1 >> E->V2 >> E->Weight;
        InsertEdge(E, Graph);
    }
    return Graph;
}

bool TopSort(LGraph Graph)
{   /* 拓扑排序 */
    int Indegree[MaxVertexNum] = {0}, cnt = 0;  // 初始化入度
    PtrToAdjNode W;
    for (int i = 0; i < Graph->Nv; i++)
    {
        for (W = Graph->G[i].FirstEdge; W; W = W->next) /* ########################## */
            Indegree[W->AdjV]++;    // 入度+1
    queue <Vertex> Q;
    for (Vertex V = 0; V < Graph->Nv; V++)
    {   /* 将入度为0的结点下标放入队列 */
        if (Indegree[V] == 0)
            Q.push(V);
    }
    while (!Q.empty())
    {
        Vertex V = Q.front();
        Q.pop();
        cnt++;  // 每出队一个元素cnt+1
        for (W = Graph->G[V].FirstEdge; W; W = W->next)
        {
            if (EarlyStart[W->AdjV] < W->Weight + EarlyStart[V])    // 更新最早开工时间
                EarlyStart[W->AdjV] = W->Weight + EarlyStart[V];
            Indegree[W->AdjV]--;    // 把所有的前驱结点为V的顶点的入度-1
            if (Indegree[W->AdjV] == 0)
                Q.push(W->AdjV);
        }
    }
    if (cnt != Graph->Nv)
        return false;
    else
        return true;
}

int main(int argc, char const *argv[])
{
    int N;
    cin >> N;
    LGraph Graph = InitialGraph(N);
    Graph = BuildGraph(Graph);
    if (!TopSort(Graph))
        cout << "Impossible" << endl;
    else
    {
        int Max = 0;
        for (int i = 0; i < N; i++) // 因为最长时间不一定在最后一个点
            if (EarlyStart[i] > Max)
                Max = EarlyStart[i];
        cout << Max << endl;
    }
    return 0;
}