假定一个工程项目由一组子任务构成，子任务之间有的可以并行执行，有的必须在完成了其它一些子任务后才能执行。“任务调度”包括一组子任务、以及每个子任务可以执行所依赖的子任务集。
比如完成一个专业的所有课程学习和毕业设计可以看成一个本科生要完成的一项工程，各门课程可以看成是子任务。有些课程可以同时开设，比如英语和C程序设计，它们没有必须先修哪门的约束；有些课程则不可以同时开设，因为它们有先后的依赖关系，比如C程序设计和数据结构两门课，必须先学习前者。
但是需要注意的是，对一组子任务，并不是任意的任务调度都是一个可行的方案。比如方案中存在“子任务A依赖于子任务B，子任务B依赖于子任务C，子任务C又依赖于子任务A”，那么这三个任务哪个都不能先执行，这就是一个不可行的方案。
任务调度问题中，如果还给出了完成每个子任务需要的时间，则我们可以算出完成整个工程需要的最短时间。在这些子任务中，有些任务即使推迟几天完成，也不会影响全局的工期；但是有些任务必须准时完成，否则整个项目的工期就要因此延误，这种任务就叫“关键活动”。
请编写程序判定一个给定的工程项目的任务调度是否可行；如果该调度方案可行，则计算完成整个工程项目需要的最短时间，并输出所有的关键活动。
输入第1行给出两个正整数N(≤100)和M，其中N是任务交接点（即衔接相互依赖的两个子任务的节点，例如：若任务2要在任务1完成后才开始，则两任务之间必有一个交接点）的数量。交接点按1~N编号，M是子任务的数量，依次编号为1~M。随后M行，每行给出了3个正整数，分别是该任务开始和完成涉及的交接点编号以及该任务所需的时间，整数间用空格分隔。
如果任务调度不可行，则输出0；否则第1行输出完成整个工程项目需要的时间，第2行开始输出所有关键活动，每个关键活动占一行，按格式“V->W”输出，其中V和W为该任务开始和完成涉及的交接点编号。关键活动输出的顺序规则是：任务开始的交接点编号小者优先，起点编号相同时，与输入时任务的顺序相反。
这个题目是How Long Does It Take 的升级版，在它的基础上得出latest[]数组，latest[v]=min{latest[w]-weight<v,w>}
然后通过delay = latest[w]-earliest[v]-weight<v,w> 如果delay = 0,那么为关键路径。
为了得到latest[]数组，我一开始使用对earliest[]的最大堆来实现的，每次将最大的earliest，拿出来。
由于我当时对latest[]更新错误，错误地认为最大堆完成不了。在这里分析一下，对于一个顶点，我们要求出它地latest，需要知道它的所有邻接点的latest，而它的邻接点的latest一定比它大，故已经算出。所以可以通过最大堆得出。
后来我用了出度来得到latest，就像通过入度得到earlist一样。每次将出度为0的点，入队，处理结束后将指向它的顶点减一后，如果出度为0就入队。
我在找错误的过程中，借鉴了别人的测试点。https://blog.csdn.net/m0_38015368/article/details/78474139
9 11 1 2 6 1 3 4 1 4 5 2 5 1 3 5 1 4 6 2 5 7 9 5 8 7 6 8 4 7 9 2 8 9 4 18 1->2 2->5 5->8 5->7 7->9 8->911 14 1 2 4 1 3 3 2 4 5 3 4 3 4 5 1 4 6 6 5 7 5 6 7 2 8 3 7 9 3 7 9 10 6 4 10 2 10 6 5 6 11 4 21 3->4 4->10 6->11 8->3 9->3 10->6
方法一：通过出度
/* How Long Does It Take*/ /*判断是否为拓扑排序，即是否存在回路*/ /*邻接表，有向图*/ /*对于得到，latest数组，因定义一个出度数组，每次将出度为0的点入队，计算他的latest，然后将新的出度为0的入队*/ /*注意此题从1开始编号*/ #include<stdio.h> #include<stdlib.h> /*边的定义*/ typedef struct Enode* Edge; struct Enode { int V1, V2; int weight; }; /*邻接点的定义*/ typedef struct Adjvnode* prttoadjvnode; struct Adjvnode { int adjv; int weight; int prt; prttoadjvnode next; }; /*顶点定义*/ typedef struct Vnode { prttoadjvnode firstedge; }adjlist[103]; /*图节点定义*/ struct Gnode { int Nv;/*顶点数*/ int Ne;/*边数*/ adjlist G; /*邻接表*/ }; typedef struct Gnode* Lgraph; /*队列定义*/ typedef struct Qnode* ptrtoQnode; struct Qnode { int* Data;/*data数组，*/ int front, rear;/*队列头尾指针*/ int maxsize;/*容量*/ }; typedef ptrtoQnode Queue; int AddQ(Queue Q, int x); Queue CreateQueue(int maxsize); int isempty(Queue Q); int DeleteQ(Queue Q); Lgraph buildgraph(int N, int M); void insertedge(Lgraph graph, Edge E); Lgraph creatgraph(int size); int main() { int N,M; scanf("%d %d", &N, &M); Lgraph graph = buildgraph(N+1, M); int* indegree = (int*)malloc(graph->Nv * sizeof(int)); int* earliest = (int*)malloc(graph->Nv * sizeof(int)); int* latest = (int*)malloc(graph->Nv * sizeof(int)); int* delay = (int*)malloc(graph->Nv * sizeof(int)); int* ondegree = (int*)malloc(graph->Nv * sizeof(int)); int v; prttoadjvnode w, j; Queue Q = CreateQueue(graph->Nv); for (v = 1; v < graph->Nv; v++) { /*初始化为0*/ earliest[v] = 0; } for (v = 1; v < graph->Nv; v++) { /*初始化为0*/ indegree[v] = 0; } for (v = 1; v < graph->Nv; v++) { /*遍历图获得入度*/ for (w = graph->G[v].firstedge; w; w = w->next) { indegree[w->adjv]++; } } for (v = 1; v < graph->Nv; v++) { /*初始化为0*/ ondegree[v] = 0; } for (v = 1; v < graph->Nv; v++) { /*遍历图获得入度*/ for (w = graph->G[v].firstedge; w; w = w->next) { ondegree[v]++; } } for (v = 1; v < graph->Nv; v++) { /*入度为0 的入队*/ if (indegree[v] == 0) { AddQ(Q, v); } } int cnt = 0; while (isempty(Q) == 0) { /*得到earliest[]数组*/ int temp_weight = 0; v = DeleteQ(Q); cnt++; for (int i = 1; i < graph->Nv; i++) { /*遍历图找到 指向V的邻接点*/ for (j = graph->G[i].firstedge; j; j = j->next) { if (j->adjv == v) { if (earliest[i] + j->weight > temp_weight) { temp_weight = earliest[i] + j->weight; } } } } earliest[v] = temp_weight; for (w = graph->G[v].firstedge; w; w = w->next) {/*对v的每个邻接点*/ if (--indegree[w->adjv] == 0) { AddQ(Q, w->adjv); } } } int min = 0; int mini; for (int i = 1; i < graph->Nv; i++) { /*输出earlist中最大的值就是最快的完成时间*/ if (earliest[i] > min) { min = earliest[i]; mini = i; } } if (cnt != graph->Nv - 1) { printf("0"); return 0; } else { printf("%d\n", min); /*下面求latest[]数组，方法与earliest一样，不过是倒着将出度为0的入队列*/ Queue Q1 = CreateQueue(graph->Nv); for (v = 1; v < graph->Nv; v++) { /*出度为0 的入队*/ if (ondegree[v] == 0) { AddQ(Q1, v); } } int v1; prttoadjvnode w2, w3; while (isempty(Q1) == 0) { v1 = DeleteQ(Q1); int temp = 200000; if(!graph->G[v1].firstedge){ temp = min; /*这里是我错的最关键的地方，我之前一直将此处写成，对于第一次出度为0的点 （即所有的终点）， 将它初始化为对应earliest[]中的值，这是不正确的，因为对于终点的最latest值，应是最大的earliest */ printf("%d ",latest[v1]); } else{ for (w2 = graph->G[v1].firstedge; w2; w2 = w2->next) { /*对于v1的每个邻接点*/ if (latest[w2->adjv] - w2->weight < temp) { temp = latest[w2->adjv] - w2->weight; } } } latest[v1] = temp; printf(" %d ",latest[v1]); for (int i = 1; i < graph->Nv; i++) { for (w3 = graph->G[i].firstedge; w3; w3 = w3->next) { /*对于指向v1的点入队*/ if (w3->adjv == v1) { if(--ondegree[i]==0){ AddQ(Q1,i); } } } } printf("\n"); } /*下面来算关键路径*/ for (int i = 1; i < graph->Nv; i++) { for (prttoadjvnode w4 = graph->G[i].firstedge; w4; w4 = w4->next) { if ((latest[w4->adjv] - earliest[i] - w4->weight) == 0) { printf("%d->%d\n", i, w4->adjv); } } } } return 0; } Lgraph creatgraph(int size) { Lgraph graph = (Lgraph)malloc(sizeof(struct Gnode)); graph->Nv = size; graph->Ne = 0; for (int i = 1; i < size; i++) { graph->G[i].firstedge = NULL; } return graph; } void insertedge(Lgraph graph, Edge E) { /*有向图*/ /*v1,v2*/ prttoadjvnode newnode = (prttoadjvnode)malloc(sizeof(struct Adjvnode)); newnode->adjv = E->V2; newnode->weight = E->weight; newnode->prt = 0; newnode->next = graph->G[E->V1].firstedge; graph->G[E->V1].firstedge = newnode; } Lgraph buildgraph(int N, int M) { Lgraph graph; graph = creatgraph(N); graph->Ne = M; if (graph->Ne != 0) { Edge E = (Edge)malloc(sizeof(struct Enode)); for (int i = 0; i < graph->Ne; i++) { scanf("%d %d %d", &E->V1, &E->V2, &E->weight); insertedge(graph, E); } } return graph; } Queue CreateQueue(int maxsize) { Queue Q = (Queue)malloc(sizeof(struct Qnode)); Q->Data = (int*)malloc(maxsize * sizeof(int)); Q->front = Q->rear = 0; Q->maxsize = maxsize; return Q; } int AddQ(Queue Q, int x) { Q->rear = (Q->rear + 1) % (Q->maxsize); /*若1%1，则值为0*/ Q->Data[Q->rear] = x; return 1; } int isempty(Queue Q) { return(Q->rear == Q->front); } int DeleteQ(Queue Q) { Q->front = (Q->front + 1) % Q->maxsize; return Q->Data[Q->front]; }方法二：最大堆
/* How Long Does It Take*/ /*判断是否为拓扑排序，即是否存在回路*/ /*邻接表，有向图*/ #include<stdio.h> #include<stdlib.h> struct data_early { int num; int value; }; typedef struct data_early array; /*堆定义*/ struct heapnode { array * data; /*存储数据的数组*/ int size; /*当前元素个数*/ int capacity; /*容量*/ }; typedef struct heapnode* Heap; /*边的定义*/ typedef struct Enode* Edge; struct Enode { int V1, V2; int weight; }; /*邻接点的定义*/ typedef struct Adjvnode* prttoadjvnode; struct Adjvnode { int adjv; int weight; int prt; prttoadjvnode next; }; /*顶点定义*/ typedef struct Vnode { prttoadjvnode firstedge; }adjlist[103]; /*图节点定义*/ struct Gnode { int Nv;/*顶点数*/ int Ne;/*边数*/ adjlist G; /*邻接表*/ }; typedef struct Gnode* Lgraph; /*队列定义*/ typedef struct Qnode* ptrtoQnode; struct Qnode { int* Data;/*data数组，*/ int front, rear;/*队列头尾指针*/ int maxsize;/*容量*/ }; typedef ptrtoQnode Queue; int AddQ(Queue Q, int x); Queue CreateQueue(int maxsize); int isempty(Queue Q); int DeleteQ(Queue Q); Lgraph buildgraph(int N, int M); void insertedge(Lgraph graph, Edge E); Lgraph creatgraph(int size); void percdown(Heap H, int p); void buildheap(Heap H); int isempty_h(Heap H); int deletemax(Heap H); int main() { int N,M; scanf("%d %d", &N, &M); Lgraph graph = buildgraph(N+1, M); //int flag = Topsort(graph); int* indegree = (int*)malloc(graph->Nv * sizeof(int)); int* earliest = (int*)malloc(graph->Nv * sizeof(int)); int* latest = (int*)malloc(graph->Nv * sizeof(int)); int* delay = (int*)malloc(graph->Nv * sizeof(int)); int v; prttoadjvnode w, j; Queue Q = CreateQueue(graph->Nv); for (v = 1; v < graph->Nv; v++) { /*初始化为0*/ earliest[v] = 0; } for (v = 1; v < graph->Nv; v++) { /*初始化为0*/ indegree[v] = 0; } for (v = 1; v < graph->Nv; v++) { /*遍历图获得入度*/ for (w = graph->G[v].firstedge; w; w = w->next) { indegree[w->adjv]++; } } for (v = 1; v < graph->Nv; v++) { /*入度为0 的入队*/ if (indegree[v] == 0) { AddQ(Q, v); } } int cnt = 0; while (isempty(Q) == 0) { int temp_weight = 0; v = DeleteQ(Q); cnt++; for (int i = 1; i < graph->Nv; i++) { /*遍历图找到 指向V的邻接点*/ for (j = graph->G[i].firstedge; j; j = j->next) { if (j->adjv == v) { if (earliest[i] + j->weight > temp_weight) { temp_weight = earliest[i] + j->weight; } } } } earliest[v] = temp_weight; for (w = graph->G[v].firstedge; w; w = w->next) {/*对v的每个邻接点*/ if (--indegree[w->adjv] == 0) { AddQ(Q, w->adjv); } } }/*while结束以后v指向最后一个顶点*/ // printf("cnt is %d num is %d\n", cnt, graph->Nv); int min = 0; int mini; for (int i = 1; i < graph->Nv; i++) { if (earliest[i] > min) { min = earliest[i]; mini = i; } } if (cnt != graph->Nv - 1) { printf("0"); return 0; } else { printf("%d\n", min); /*如果下面这个失败就用最大堆来做*/ /*最大堆*/ /*创建堆*/ Heap H = (Heap)malloc(sizeof(struct heapnode)); H->data = (array*)malloc(graph->Nv * sizeof(struct data_early)); H->capacity = graph->Nv; H->size = graph->Nv - 1; H->data[0].value = 20000; for (int i = 1; i < graph->Nv; i++) { H->data[i].num = i; H->data[i].value = earliest[i]; } buildheap(H);/*修改为最大堆*/ int v1; prttoadjvnode w2; while (isempty_h(H)==0) { v1 = deletemax(H); // printf("%d is",v1); if (!graph->G[v1].firstedge) { /*如果没有邻接点*/ latest[v1] = min; // printf("%d ",latest[v1]); } else { int temp = 200000; for (w2 = graph->G[v1].firstedge; w2; w2 = w2->next) { /*对于v1的每个邻接点*/ //printf("ear is %d weight is %d ",earliest[w2->adjv],w2->weight); if (latest[w2->adjv] - w2->weight < temp) { /*错误原因表达式输入错误*/ temp = latest[w2->adjv] - w2->weight; } } latest[v1] = temp; // printf(" %d ",latest[v1]); } //printf("\n"); } /* printf("-----\n"); for (int i = 1; i < graph->Nv; i++) { printf("%d ", latest[i]); } printf("\n");*/ /*下面来算关键路径*/ for (int i = 1; i < graph->Nv; i++) { for (prttoadjvnode w4 = graph->G[i].firstedge; w4; w4 = w4->next) { if ((latest[w4->adjv] - earliest[i] - w4->weight) == 0) { printf("%d->%d\n", i, w4->adjv); } } } /* for (int i = 1; i < graph->Nv; i++) { printf("第%d： ",i); for (prttoadjvnode w4 = graph->G[i].firstedge; w4; w4 = w4->next) { printf("%d ",w4->adjv); } printf("\n") ; }*/ } return 0; } Lgraph creatgraph(int size) { Lgraph graph = (Lgraph)malloc(sizeof(struct Gnode)); graph->Nv = size; graph->Ne = 0; for (int i = 1; i < size; i++) { graph->G[i].firstedge = NULL; } return graph; } void insertedge(Lgraph graph, Edge E) { /*有向图*/ /*v1,v2*/ prttoadjvnode newnode = (prttoadjvnode)malloc(sizeof(struct Adjvnode)); newnode->adjv = E->V2; newnode->weight = E->weight; newnode->prt = 0; newnode->next = graph->G[E->V1].firstedge; graph->G[E->V1].firstedge = newnode; } Lgraph buildgraph(int N, int M) { Lgraph graph; graph = creatgraph(N); graph->Ne = M; if (graph->Ne != 0) { Edge E = (Edge)malloc(sizeof(struct Enode)); for (int i = 0; i < graph->Ne; i++) { scanf("%d %d %d", &E->V1, &E->V2, &E->weight); insertedge(graph, E); } } return graph; } Queue CreateQueue(int maxsize) { Queue Q = (Queue)malloc(sizeof(struct Qnode)); Q->Data = (int*)malloc(maxsize * sizeof(int)); Q->front = Q->rear = 0; Q->maxsize = maxsize; return Q; } int AddQ(Queue Q, int x) { Q->rear = (Q->rear + 1) % (Q->maxsize); /*若1%1，则值为0*/ Q->Data[Q->rear] = x; return 1; } int isempty(Queue Q) { return(Q->rear == Q->front); } int DeleteQ(Queue Q) { Q->front = (Q->front + 1) % Q->maxsize; return Q->Data[Q->front]; } void percdown(Heap H, int p) { int parent, child; struct data_early x; x = H->data[p]; for (parent = p; parent * 2 <= H->size; parent = child) { child = parent * 2; if ((child != H->size) && (H->data[child].value < H->data[child + 1].value)) { child++; } if (x.value >= H->data[child].value)break; else { H->data[parent] = H->data[child]; } } H->data[parent] = x; } void buildheap(Heap H) { int i = 0; for (i = H->size / 2; i > 0; i--) { percdown(H, i); } } int isempty_h(Heap H) { return(H->size == 0); } int deletemax(Heap H) { int parent, child; struct data_early x; int max = H->data[1].num; x = H->data[H->size--]; for (parent = 1; parent * 2 <= H->size; parent = child) { child = parent * 2; if ((child != H->size) && (H->data[child].value < H->data[child + 1].value)) { child++; } if (x.value >= H->data[child].value)break; else { H->data[parent] = H->data[child]; } } H->data[parent] = x; return max; }
