关键路径

了解：关键路径中边表示活动，点表示事件。关键路径主要在求得两个问题1.完成这项工程至少需要多少时间 2.那些活动时关键的

1.求出所有事件的最早发送时间(ve[N])，最迟发生时间(vl[N])

2.根据ve[]和vl[]求出活动的v[]和l[]

3.当某点v[i]=l[i]时就是关键活动

#include<iostream>
#include<string.h>
#include<algorithm>
#include<stack>
#define N 100
using namespace std;
struct graph{
    int a[N][N];
    int n;
}; 

int ve[N],vl[N];
stack<int> t;

int TopologicalSort(graph g){
    int indegree[N],i,j,count;
    stack<int> s;
    
    memset(indegree,0,sizeof(indegree));
    count=0;
    for(i=0;i<g.n;i++)
        for(j=0;j<g.n;j++)
            if(g.a[i][j]>0)indegree[j]++; 
    for(i=0;i<g.n;i++)if(!indegree[i]){s.push(i);t.push(i);}
    
    while(!s.empty()){
        i=s.top();
        s.pop();
        count++;
        for(j=0;j<g.n;j++){
            if(g.a[i][j]){
                indegree[j]--;
                if(!indegree[j]){s.push(j);t.push(j);}
                ve[j]=max(ve[i]+g.a[i][j],ve[j]);
            }
        }
    }
    if(count<g.n)return 0;
    return 1;
}

int CriticalPath(graph g){
    if(!TopologicalSort(g))return 0;
    int i,j;
    int ee,el;
    char tag;
    for(i=0;i<g.n;i++)vl[i]=ve[g.n-1];
    
    while(!t.empty()){
        i=t.top();
        t.pop();
        for(j=0;j<g.n;j++)if(g.a[i][j])vl[i]=min(vl[i],vl[j]-g.a[i][j]);
    }
    
    for(i=0;i<g.n;i++){
        for(j=0;j<g.n;j++){
            if(g.a[i][j]){
                ee=ve[i],el=vl[j]-g.a[i][j];
                tag=(ee==el)?'*':' ';
                printf("%d %d %d %d %d %c\n",i+1,j+1,g.a[i][j],ee,el,tag);
            }
        }
    }
    cout<<endl; 
    for(i=0;i<g.n;i++)cout<<ve[i]<<" ";
    cout<<endl;
    for(i=g.n-1;i>=0;i--)cout<<vl[i]<<" ";
    
    return 1;
} 


int main(){
    graph g;
    cin>>g.n;
    int i,j;
    for(i=0;i<g.n;i++){
        for(j=0;j<g.n;j++){
            cin>>g.a[i][j];
        }
    }
    CriticalPath(g);
    return 0;
}

测试数据：

9
0 6 4 5 0 0 0 0 0
0 0 0 0 1 0 0 0 0
0 0 0 0 1 0 0 0 0
0 0 0 0 0 2 0 0 0
0 0 0 0 0 0 9 7 0
0 0 0 0 0 0 0 4 0
0 0 0 0 0 0 0 0 2
0 0 0 0 0 0 0 0 4
0 0 0 0 0 0 0 0 0

邻接表＋java

import java.util.LinkedList;
import java.util.Queue;
import java.util.Scanner;
import java.util.Stack;

class Edge{
	int from;
	int to;
	int w;
	int next;
}

public class Main{
	static Edge[] edges=new Edge[10005];
	static int[] head=new int[10005];
	
	public static void add(int from,int to,int w,int i) {
		edges[i]=new Edge();
		edges[i].from=from;
		edges[i].to=to;
		edges[i].w=w;
		edges[i].next=head[from];
		head[from]=i;
	}
	
	public static void main(String[] args) {
		Queue<Integer> que1=new LinkedList<Integer>();
		Stack<Integer> que2=new Stack<Integer>();
		int[] indegree=new int[10005];
		int[] ve=new int[10005];//事件最早
		int[] vl=new int[10005];//事件最晚
		
		Scanner sc=new Scanner(System.in);
		int n,m;
		int from,to,w;
		int i,j,top;
		int ee,el;
		
		n=sc.nextInt();
		m=sc.nextInt();
		for(i=1;i<=n;i++)head[i]=-1;
		for(i=1;i<=m;i++) {
			from=sc.nextInt();
			to=sc.nextInt();
			w=sc.nextInt();
			add(from, to, w,i);
			indegree[to]++;
		}
		for(i=1;i<=n;i++) {
			if(indegree[i]==0) {
				que1.add(i);
				que2.add(i);
			}
		}
		
		while(!que1.isEmpty()) {
			top=que1.poll();
			for(i=head[top];i!=-1;i=edges[i].next) {
				if(indegree[edges[i].to]>0) {
					ve[edges[i].to]=Math.max(ve[edges[i].to],ve[top]+edges[i].w);
					indegree[edges[i].to]--;
					if(indegree[edges[i].to]==0) {
						que1.add(edges[i].to);
						que2.add(edges[i].to);
					}
				}
			}
		}
		
		for(i=1;i<=n;i++)vl[i]=ve[n];
		while(!que2.isEmpty()) {
			top=que2.pop();
			for(i=1;i<=n;i++) {
				for(j=head[i];j!=-1;j=edges[j].next) {
					if(edges[j].to==top) {
						vl[edges[j].from]=Math.min(vl[edges[j].from],vl[edges[j].to]-edges[j].w);
					}
				}
			}
		}
		
		for(i=1;i<=n;i++) {
			for(j=head[i];j!=-1;j=edges[j].next) {
				ee=ve[i];
				el=vl[edges[j].to]-edges[j].w;
				char tag=ee==el?'*':'X';
				System.out.println(i+" "+edges[j].to+" "+ee+" "+el+tag);
			}
		}
	}
}

9 11
1 2 6
1 3 4
1 4 5
2 5 1
3 5 1
4 6 2
5 7 9
5 8 7
6 8 4
7 9 2
8 9 4

关键路径

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