数据结构基础— How Long Does It Take

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 $N$ (≤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

//
//  main.cpp
//  TopSort
//
//  Created by air on 2018/4/26.
//  Copyright © 2018年 air. All rights reserved.
/*这道题是 Earliest[j] = max{Earliest[i] + C(i,j)};
 利用拓扑排序进行改造
 即使下一个邻接点的入度不为0，一样需要更新它的Earliest的值，只是不会入队，只有在入度为0的时候才可以入队
*/
#include 
#include 
#include 
#include 
#include 
#define Infinity 1 << 30
using namespace std;
typedef int vertex;
typedef int degree;
typedef int weightType;
struct GNode{
    vertex Nv;
    vertex Ne;
    int** map;
};
typedef struct GNode* ptrToGraph;
typedef ptrToGraph LGraph;
/*---------------------建立图----------------------*/
LGraph buildGraph(){
    LGraph graph = new GNode;
    cin >> graph->Nv >> graph->Ne;
    /* 建立二维数组 */
    graph->map = new vertex*[graph->Nv];
    for(vertex i = 0; i < graph->Nv; i++)
        graph->map[i] = new vertex[graph->Nv];
    
    /* 一开始初始化成正无穷大*/
    for(int i = 0; i < graph->Nv; i++){
        for(int j = 0; j < graph->Nv; j++)
            graph->map[i][j] = Infinity;
    }
    /* 数值初始化,有向图 */
    vertex V1,V2;
    weightType Weight;
    for(int i = 0; i < graph->Ne; i++){
        cin >> V1 >> V2 >> Weight;
        graph->map[V1][V2] = Weight;
    }
    return graph;
}
/*---------------------拓扑排序----------------------*/
void TopSort(LGraph graph){
    vertex i,j;
    int cnt = 0;
    queue q;
    int Indegree[graph->Nv];
    int Earliest[graph->Nv];
    
    for(i = 0; i < graph->Nv; i++){
        Indegree[i] = 0;
        Earliest[i] = 0;
    }
    
    for(i = 0; i < graph->Nv; i++){
        for(j = 0; j < graph->Nv; j++){
            if(graph->map[i][j] != Infinity)
                Indegree[j] ++;
        }
    }
    /* 一开始把入度为0的全部入队列 */
    for(i = 0; i < graph->Nv; i++)
        if(Indegree[i] == 0)
            q.push(i);

    while (! q.empty()) {
        vertex temp = q.front();
        q.pop();
        cnt++;
        for(i = 0; i < graph->Nv; i++){
            if(graph->map[temp][i] != Infinity){
                Earliest[i] = max(Earliest[i], Earliest[temp] + graph->map[temp][i]);
                if((--Indegree[i]) == 0)
                /* 这个if一定要在上一个if的下面，我一开始的错就在放一起了，所以导致一个测试点一直过不了 */
                    q.push(i); /* 入度为0才入队，并不是直接临接点就入队 */
            }
        }
    }
    
    /* 以下解决了多个起点和多个终点第四个评测点 */
    int maxNumber = 0;
    for(i = 0; i < graph->Nv; i++)
        maxNumber = max(maxNumber,Earliest[i]);
    
    if (cnt != graph->Nv)
        cout << "Impossible" << endl;
    else
        cout << maxNumber << endl;
}

int main(int argc, const char * argv[]) {
    LGraph graph = buildGraph();
    TopSort(graph);
    delete graph;
    return 0;
}