《数据结构》08-图8 How Long Does It Take

题目

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

分析

考察拓扑排序
每次更新一个点时，更新其周围邻接点的时间
最后输出时，也许有多个终点，输出时间最大的，即为项目最早完成时间

#include<iostream>
#include<queue>
#include <algorithm>
#define MaxVertex 105
#define INF -100000
typedef int Vertex;
using namespace std;
int N; // 点
int M;  // 边
int G[MaxVertex][MaxVertex];  
int Earliest[MaxVertex];  // 时间 
int Indegree[MaxVertex]; // 入度 
// 初始化图 
void build(){
    Vertex v1,v2,w;
    cin>>N>>M;
    for(Vertex i=0;i<N;i++){
        for(Vertex j=0;j<N;j++)
            G[i][j] = INF;
    }
    for(int i=0;i<M;i++){
        cin>>v1>>v2>>w;
        G[v1][v2] = w;  // 有向图 
        Indegree[v2]++;  // 入度+1 
    }
}
void TopSort(){
    int cnt = 0;
    queue<Vertex> q;
    // 入度为0顶点入队 
    for(Vertex i=0;i<N;i++)
        if(!Indegree[i]){
            q.push(i);
            Earliest[i] = 0;
        } 
    while(!q.empty()){
        Vertex v = q.front();
        q.pop();
        cnt++;
        for(Vertex w=0;w<N;w++)
            if(G[v][w]!=INF){
                if(Earliest[w] < Earliest[v]+G[v][w])  //如果周围有时间更长，更新时间 
                    Earliest[w] = max(Earliest[w],Earliest[v]+G[v][w]);
                if(--Indegree[w]==0)
                    q.push(w);
            } 
    }
    if(cnt!=N)
        cout<<"Impossible";
    else{
        // 也许不止一个终点 
        int max = 0;
        for(Vertex i=0;i<N;i++)  
            if(max < Earliest[i]) 
                max = Earliest[i];
        cout<<max;
    }
}
int main(){
    build();
    TopSort();
    return 0;
}