题目描述
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, 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
解题思路
本题是拓扑排序的变形,需要一个数组来记录到顶点的距离,在拓扑排序的过程中只需要不断更新该数组即可。
代码
#include <stdio.h>
#include <stdlib.h>
#define MAXSIZE 100
struct ENode {
int target;
int weight;
struct ENode *next;
};
typedef struct ENode *Edge;
typedef struct {
Edge fisrtEdge;
} List[MAXSIZE];
struct Graph {
int vertexCount;
int edgeCount;
List header;
};
typedef struct Graph *LGraph;
LGraph createGraph();
void topSort(LGraph graph);
int main() {
LGraph graph = createGraph();
topSort(graph);
return 0;
}
LGraph createGraph() {
int N, M;
scanf("%d %d", &N, &M);
LGraph graph = (LGraph) malloc(sizeof(struct Graph));
graph->vertexCount = N;
graph->edgeCount = M;
for (int i = 0; i < N; i++)
graph->header[i].fisrtEdge = NULL;
for (int i = 0; i < M; i++) {
int x, y, weight;
scanf("%d %d %d", &x, &y, &weight);
Edge newEdge = (Edge) malloc(sizeof(struct ENode));
newEdge->target = y;
newEdge->weight = weight;
newEdge->next = graph->header[x].fisrtEdge;
graph->header[x].fisrtEdge = newEdge;
}
return graph;
}
void topSort(LGraph graph) {
int counter = 0; //判断是否有回路
int dis[MAXSIZE] = {0}; //记录到每个顶点的距离
int indegree[MAXSIZE] = {0}; //记录每个顶点的入度
int queue[MAXSIZE], front = 0, rear = 0; //队列
for (int i = 0; i < graph->vertexCount; i++)
for (Edge e = graph->header[i].fisrtEdge; e; e = e->next)
indegree[e->target]++; //初始化入度
for (int i = 0; i < graph->vertexCount; i++)
if (indegree[i] == 0)
queue[rear++] = i; //将入度为0的顶点入队
while (front != rear) { //拓扑排序核心代码
int vertex = queue[front++];
counter++;
for (Edge e = graph->header[vertex].fisrtEdge; e; e = e->next) {
if (dis[vertex] + e->weight > dis[e->target])
dis[e->target] = dis[vertex] + e->weight; //更新dis数组
if (--indegree[e->target] == 0) queue[rear++] = e->target;
}
}
if (counter == graph->vertexCount) { //判断是否有回路
int max = 0;
for (int i = 0; i < counter; i++)
if (dis[i] > max) max = dis[i];
printf("%d
", max);
} else {
printf("Impossible
");
}
}