题目:
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 4Sample Output 1:
18Sample Input 2:
4 5 0 1 1 0 2 2 2 1 3 1 3 4 3 2 5Sample Output 2:
Impossible
胡乱分析:
一道挺典型的拓扑排序,先把所有入度为0的点入列,进入循环,每次弹出一个节点,找出与这个节点相连的节点,入度减1,更新最小花费时间,再把所有入度为0的节点入列,进入下一次循环。加了个cnt判断有没有环。
,代码:
#include <iostream> #include <algorithm> #include <queue> #define MAXN 105 #define INF 1e7 using namespace std; int Graph[MAXN][MAXN];//矩阵Graph[i][j]代表i->j的cost void TopSort(int N) { int cnt = 0; //计数器判断有无环路 int* Earliest = new int[N]; //节点的时间花费 int* Indegree = new int[N]; //入度 for (int i = 0; i < N; i++) { Indegree[i] = 0; Earliest[i] = 0; for (int j = 0; j < N; j++) { //初始化 入度 if (Graph[j][i] != INF) { Indegree[i]++; } } } queue<int> Q; for (int i = 0; i < N; i++) { //起始判断入度为0的节点,放入队列(其实这里用堆栈也可以 if (Indegree[i] == 0) { Q.push(i); Indegree[i] = -1; //标记节点i避免重复入队 Earliest[i] = 0; } } while (!Q.empty()) { int temp = Q.front(); Q.pop(); cnt++; //处理了一个节点cnt++ int longest = 0; for (int i = 0; i < N; i++) { //遍历所有节点 if (Graph[temp][i] != INF) { //如果有边<temp,i> Indegree[i]--; //i的的入度减1 Earliest[i] = max(Earliest[i], Earliest[temp] + Graph[temp][i]);//节点i的最长花费进行更新 } } for (int i = 0; i < N; i++) {//再次循环判断有没有入度为0的节点,如果有就放入队列 if (Indegree[i] == 0) { Q.push(i); Indegree[i] = -1; //标记节点避免重复入队 } } } sort(Earliest, Earliest + N); //对所有节点的时长花费进行从小到大的排序,也处理了有多个终点的情况 if (cnt == N) { //如果处理了N个节点 cout << Earliest[N - 1]; //输出最长的时间花费 } else { cout << "Impossible"; //否则代表图中有环 } } int main() { int N, M, v1, v2, cost;//节点个数N 边数M cin >> N >> M; for (int i = 0; i < N; i++) { //初始化整张图INF for (int j = 0; j < N; j++) { Graph[i][j] = INF; } } for (int i = 0; i < M; i++) { //输入<v1,v2>的cost cin >> v1 >> v2 >> cost; Graph[v1][v2] = cost; } TopSort(N); }