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
思路：
关键路径问题，以 dist[i] 数组记录到 i 节点的最短时间。以 inDegree[i] 数组记录 i 节点的入度。
每次把入度为零的节点入队，入队之后将该节点指向的节点入度减一。
#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<queue>
using namespace std;
const int N = 102;
queue<int> q;
int graph[N][N] = { 0 };
int inDegree[N] = { 0 };
int dist[N];
bool topsort(int n) {
//初始化dist
for (int i = 0; i < n; i++) {
dist[i] = -1;
}
int cur = 0;
for (int i = 0; i < n; i++) {
if (inDegree[i] == 0) {
q.push(i);
dist[i] = 0;
for (int j = 0; j < n; j++) {
if (graph[i][j] > 0) {
dist[j] = graph[i][j];
}
}
}
}
int cnt = 0;
while (!q.empty()) {
cur = q.front();
q.pop();
cnt++;
for (int i = 0; i < n; i++) {
if (graph[cur][i] >= 0) {
inDegree[i]--;
if (dist[cur] + graph[cur][i] > dist[i]) {
dist[i] = graph[cur][i] + dist[cur];
}
if (inDegree[i] == 0) {
q.push(i);
}
}
}
}
if (cnt != n) {
return false;
}
return true;
}
int main() {
int n, m;
cin >> n >> m;
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
graph[i][j] = -1;
}
}
while (m--) {
int s, e, l;
cin >> s >> e >> l;
graph[s][e] = l;
inDegree[e]++;
}
if (!topsort(n))
cout << "Impossible" << endl;
else {
int max = -1;
for (int i = 0; i < n; i++) {
max = max > dist[i] ? max : dist[i];
}
cout << max << endl;
}
return 0;
}