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 $N$ ($\le 100$), 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:

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 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; }