How Long Does It Take (25分)【C语言】AOE拓扑排序_how long does it take用c语言-优快云博客

本文介绍如何使用AOE算法与拓扑排序解决项目调度问题，详细解析了输入输出格式，通过实例展示了如何计算项目的最早完成时间，并提供了代码实现。

题目：

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”.

输入样例

输出样例

算法

AOE拓扑排序

入度为0的结点全部压入队列（如果是单一起点，压入指点结点即可）
while循环中弹出一个结点，对此节点的邻结点的入度减一（如果此邻结点入度减为0则直接入队）同时，更新此结点的earliest
根据处理的结点总数判断时间安排是否合理。合理时，在所有结点中选择最大的earliest输出

代码实现

int main()
{	
	int N,E;
	scanf("%d %d",&N,&E);
	Graph G=CreateGraph(N);
	BuildGraph(G,E);
	AOE(G);
	return 0;
}

AOE函数

入度为0的结点全部压入队列（如果是单一起点，压入指点结点即可）
while循环中弹出一个结点，对此节点的邻结点的入度减一（如果此邻结点入度减为0则直接入队）同时，更新此结点的earliest
根据处理的结点总数判断时间安排是否合理。合理时，在所有结点中选择最大的earliest输出

void AOE(Graph G)
{
	int *earliest=(int*)malloc(sizeof(int)*(G->VertexNum));//记录每个结点完成的最早时间 
	int *indegree=(int*)malloc(sizeof(int)*(G->VertexNum));//记录每个结点的入度 
	int i,j;
	for(i=0;i<G->VertexNum;i++){//初始化earliest，indegree
	    earliest[i]=-1;
	    indegree[i]=0;
	    for(j=0;j<G->VertexNum;j++){
	        if(G->GraphMatrix[j][i]!=-1){
	            indegree[i]++;
	        }
	    }
	}
	int *Queue=(int*)malloc(sizeof(int)*(G->VertexNum+1));
	int rear=0,head=0;
	int count=0;//当前收录到集合中元素个数 
	for(i=0;i<G->VertexNum;i++){//收录初始入度为0的结点
	    if(indegree[i]==0){ 
	        Queue[rear++]=i;
	        earliest[i]=0;
	    }
	}
	int t,MaxVertex;
	while(rear>head){
	    t=Queue[head++];
	    count++;
	    for(i=0;i<G->VertexNum;i++){
	        if((G->GraphMatrix[t][i]!=-1)){
	            if(--indegree[i]==0){
	                Queue[rear++]=i;
	            }
	            if(earliest[i]<earliest[t]+G->GraphMatrix[t][i]){
	                earliest[i]=earliest[t]+G->GraphMatrix[t][i];
	            } 
	        }
	    } 
	}
	if(count==G->VertexNum){
	    int MaxSchedule=0;
	    for(i=0;i<G->VertexNum;i++){
	        if(earliest[i]>MaxSchedule){
	            MaxSchedule=earliest[i];
	        }
	    }
	    printf("%d",MaxSchedule);
	}else{
	    printf("Impossible");
	}
}

邻接矩阵存储的图

#define MAXVERTEXNUM 101
typedef struct GNode* Graph;
struct GNode{
	int VertexNum;
	int EdgeNum;
	int GraphMatrix[MAXVERTEXNUM][MAXVERTEXNUM];
};
typedef struct ENode* edge;
struct ENode{
	int V;
	int W;
	int Weight;
};
Graph CreateGraph(int N)
{
	Graph G=(Graph)malloc(sizeof(struct GNode));
	G->VertexNum=N;
	int i,j;
	for(i=0;i<N;i++){
	    for(j=0;j<N;j++){
	        G->GraphMatrix[i][j]=-1;
	    }
	}
	return G;
}
void InsertEdge(Graph G,edge L)
{
	G->GraphMatrix[L->V][L->W]=L->Weight;
}
void BuildGraph(Graph G,int E)
{
	G->EdgeNum=E;
	edge L=(edge)malloc(sizeof(struct ENode));
	int i;
	for(i=0;i<E;i++){
	    scanf("%d %d %d",&(L->V),&(L->W),&(L->Weight));
	    InsertEdge(G,L);
	}
	free(L);
}