标签:sam node com ram esc efi sizeof sem tom
Given the relations of all the activities of a project, you are supposed to find the earliest completion time of the project.
Each input file contains one test case. Each case starts with a line containing two positive integers N (≤), 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.
For each test case, if the scheduling is possible, print in a line its earliest completion time; or simply output "Impossible".
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
18
4 5
0 1 1
0 2 2
2 1 3
1 3 4
3 2 5
Impossible
1 //拓扑排序 2 #include<stdio.h> 3 #include<stdlib.h> 4 #define Max 105 5 #define ERROR -1 6 7 int N, M; 8 int Graph[Max][Max]; //邻接矩阵储存图 9 int Indegree[Max] = {0}; //记录每个结点的入度 10 int time[Max] = {0}; //记录每个结点需要花费的时间 11 12 void init(){ 13 for(int i=0; i<N; i++){ 14 for(int j=0; j<N; j++){ 15 Graph[i][j] = ERROR; 16 } 17 } 18 } 19 20 void BuildG(){ 21 scanf("%d%d", &N, &M); 22 init(); 23 for(int i=0; i<M; i++){ 24 int S, E, time; 25 scanf("%d%d%d", &S, &E, &time); 26 Graph[S][E] = time; 27 Indegree[E]++; 28 } 29 } 30 31 typedef struct QNode{ 32 int *data; 33 int rear, front; 34 int maxsize; 35 }*Queue; 36 37 Queue BuildQ(){ 38 Queue Q = (Queue)malloc(sizeof(struct QNode)); 39 Q->data = (int *)malloc(sizeof(int)*N); 40 Q->front = Q->rear = 0; 41 Q->maxsize = N; 42 return Q; 43 } 44 45 void AddQ(Queue Q, int x){ 46 Q->rear = (Q->rear + 1)%Q->maxsize; 47 Q->data[Q->rear] = x; 48 } 49 50 bool IsEmpty(Queue Q){ 51 return Q->front == Q->rear; 52 } 53 54 int DeleteQ(Queue Q){ 55 Q->front = (Q->front + 1)%Q->maxsize; 56 return Q->data[Q->front]; 57 } 58 59 void TopSort(){ 60 Queue Q = BuildQ(); //用队列记录入度为0的结点 61 for(int i=0; i<N; i++){ 62 if(Indegree[i]==0){ 63 AddQ(Q, i); 64 } 65 } 66 int v, cnt=0, Time; 67 while(!IsEmpty(Q)){ 68 v = DeleteQ(Q); 69 cnt++; 70 for(int i=0; i<N; i++){ 71 if(Graph[v][i]!=ERROR){ 72 Indegree[i]--; 73 if(Indegree[i]==0){ 74 AddQ(Q, i); 75 } 76 if(time[v] + Graph[v][i] > time[i]){ 77 time[i] = time[v] + Graph[v][i]; 78 } 79 if(time[i]>Time){ 80 Time = time[i]; 81 } 82 } 83 } 84 } 85 if(cnt<N){ //如果有几个结点不能被出队,则一定存在闭环,输出impossible 86 printf("Impossible\n"); 87 } 88 else printf("%d\n", Time); 89 } 90 91 int main(){ 92 BuildG(); 93 TopSort(); 94 return 0; 95 }
标签:sam node com ram esc efi sizeof sem tom
原文地址:https://www.cnblogs.com/shin0324/p/9978716.html