7-12 How Long Does It Take（25 分）

最新推荐文章于 2019-08-31 20:37:46 发布

Allenhong97

最新推荐文章于 2019-08-31 20:37:46 发布

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分类专栏： PAT 文章标签： PAT 关键路径拓扑排序

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本文介绍了一个项目调度算法的具体实现，该算法能够根据项目的活动关系找出项目的最早完成时间。输入包括活动检查点数量、活动数量及各活动的起始点、终点和持续时间。通过构建图模型并采用拓扑排序的方法来解决此问题。

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7-12 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 $N$ ( $\leq 100$ ), 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:

Sample Output 1:

Sample Input 2:

Sample Output 2:

Impossible

#include<iostream>
#include<vector>
#include<queue>
#include<stdio.h>
#include<utility>
using namespace std;

int Nv,Ne;

vector<int>G[105];

struct Node{
	int in;//入度
	int earliest;//最早完成时间
	vector<pair<int,int>>V;//pair<to,weight>，存放边
}N[105];

void Read(){
	int a;
	int b;
	int c;
	for(int i=0;i<Nv;i++){
		N[i].in=0;
		N[i].earliest=0;
	}
	for(int i=0;i<Ne;i++){
		cin>>a>>b>>c;
		N[b].in++;
		G[a].push_back(b);
		G[b].push_back(a);
		N[a].V.push_back(make_pair(b,c));
	}
}

void TopSort(){
	int cnt=0;
	int LeastTime=0;
	queue<int>Q;
	for(int i=0;i<Nv;i++){
		if(N[i].in==0)
			Q.push(i);
	}
	while (Q.size()>0){
		int temp=Q.front();
		Q.pop();
		cnt++;
		int lenv=N[temp].V.size();
		for(int i=0;i<lenv;i++){
			int to=N[temp].V[i].first;
			int weight=N[temp].V[i].second;
			N[to].in--;
			if(N[to].in==0) Q.push(to);
			N[to].earliest=max(N[to].earliest,N[temp].earliest+weight);
			LeastTime=max(N[to].earliest,LeastTime);
		}
	}
	if(cnt==Nv)
		cout<<LeastTime<<endl;
	else
		cout<<"Impossible"<<endl;
}

void Print(){
	for(int i=0;i<Nv;i++){
		cout<<"Node: "<<i<<"Time: "<<N[i].earliest<<endl;
	}
}

int main(){
	//freopen("in1.txt","r",stdin);
	cin>>Nv>>Ne;
	Read();
	TopSort();
	//Print();
	//fclose(stdin);
	return 0;
}