陈越何欣铭老师数据结构PTA08-图8 How Long Does It Take

题目：

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 NNN (≤100\le 100≤100), the number of activity check points (hence it is assumed that the check points are numbered from 0 to N−1N-1N−1), and MMM, the number of activities. Then MMM 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

思路：只可意会不可言传。基本和陈越老师讲的例题意思一样。

代码：

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
#define MaxN 100+10
typedef struct line *queue;
int indegree[MaxN]={0};
int maxtime[MaxN];
int symbol[MaxN]={0};
struct act
{int *outact;int *outtime;int now;
}a[MaxN];
struct line
{int star;int end;int *Element;
};
//
queue CreateQ(int Size)
{queue q;q=(queue)malloc(sizeof(struct line));q->star=0;q->end=0;q->Element=(int *)malloc(sizeof(int)*(Size+1));return q;
}
void InsertQ(int X,queue q)
{q->Element[q->end++]=X;
}
int DeleteQ(queue q)
{return q->Element[q->star++];
}
int IsEmptyQ(queue q)
{return q->star==q->end;
}
//
void Initialize(int N)
{int i;for(i=0;i<=N;i++){a[i].now=0;a[i].outact=(int *)malloc(sizeof(int)*(N+1));a[i].outtime=(int *)malloc(sizeof(int)*(N+1));}
}
int main()
{int N,M;queue q;memset(maxtime,0,sizeof(maxtime));scanf("%d%d",&N,&M);Initialize(N);q=CreateQ(N);int i;for(i=1;i<=M;i++){int star,end,time;scanf("%d%d%d",&star,&end,&time);a[star].outact[a[star].now]=end;a[star].outtime[a[star].now++]=time;indegree[end]++;}for(i=0;i<N;i++)if(indegree[i]==0){InsertQ(i,q);symbol[i]=1;}int cnt=0;int X=-1; while(!IsEmptyQ(q)){X=DeleteQ(q);cnt++;for(i=0;i<a[X].now;i++){indegree[a[X].outact[i]]--;if(indegree[a[X].outact[i]]==0)InsertQ(a[X].outact[i],q);if(maxtime[a[X].outact[i]]<maxtime[X]+a[X].outtime[i])maxtime[a[X].outact[i]]=maxtime[X]+a[X].outtime[i];} }if(cnt!=N)printf("Impossible");else{int max=-1;for(i=0;i<N;i++)if(maxtime[i]>max)max=maxtime[i];  printf("%d",max);} return 0;
}