08-图8 How Long Does It Take (25分)

简单的拓扑排序。

基础思路是这样的，对每个节点，都有一个时间参数，用来记录从起始到这个节点总共所需的最短时间。由于规矩是只有把该节点所有的入度给做完了，才能执行这个节点。所以每个节点都是等于它的入度节点的以及路径权重相加的最大值。

利用拓扑排序算法就可以做了。注意多起始节点和多终止节点怎么处理的问题。

#include 
#include 

typedef struct _node{
	int name;
	int time;
	struct _node *next;
}Node;

typedef struct _head{
	int in;
	struct _node *out;
}Head;

typedef struct _que{
	int loction;
	struct _que *next;
}Que;

typedef struct _table{
	int know;
	int stime;
}Table;

Table *Sheet;
Head *ArrayList;
int Check;
int Act;
int First;
int Last;

void add_array(Node *Beginer,int name,int time);
void print_array(Node *Beginer);

void insert_que(Que *root,int i);
int  dele_que(Que *root);
void topshot();
void print_table();
void print_que(Que *root);

int main(){
	scanf("%d %d",&Check,&Act);
	ArrayList=(Head*)malloc(Check*sizeof(Head));
	Sheet=(Table*)malloc(Check*sizeof(Table));
	int i,j,t,k;
	for(k=0;knext=NULL;
		Sheet[k].know=0;
		Sheet[k].stime=0;
	}
	
	for(k=0;knext=NULL;
	int k;
	int cnt=0;
	for(k=0;knext;
		while(subhead){
			if(Sheet[subhead->name].know==0){
				--ArrayList[subhead->name].in;
				if(ArrayList[subhead->name].in==0){
					insert_que(root,subhead->name);
					cnt++;
				}
				if(Sheet[result].stime+subhead->time > Sheet[subhead->name].stime){
					Sheet[subhead->name].stime=Sheet[result].stime+subhead->time;
				}	
			}
			subhead=subhead->next;
		}
		
	}
	if(cnt!= Check){
		printf("Impossible");
	}else{
		int biggest=0;
		for(k=0;kbiggest){
				biggest=Sheet[k].stime;
			}
		}
		printf("%d",biggest);
	}
}

void add_array(Node *Beginer,int name,int time){
	Node *p=Beginer;
	Node *q=(Node*)malloc(sizeof(Node));
	q->name=name;
	q->time=time;
	q->next=Beginer->next;
	Beginer->next=q;
}

void print_array(Node *Beginer){
	Node *p=Beginer->next;
	while(p){
		printf("name=%d time=%d% ",p->name,p->time);
		p=p->next;
	}
	
}

void insert_que(Que *root,int i){
	Que *p=root;
	while(p->next){
		p=p->next;
	}
	Que *q=(Que*)malloc(sizeof(Que));
	q->loction=i;
	p->next=q;
	q->next=NULL;
}

int  dele_que(Que *root){
	Que *p=root->next;
	int result=-1;
	if(p){
		result=p->loction;
		root->next=p->next;
		free(p);
	}
	return result;
}

void print_table(){
	int k=0;
	for(k=0;knext;
	while(p){
		printf("%d ",p->loction);
		p=p->next;
	}
	printf("\n");
}

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, andL[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