[최단 경로] 2171039 이채원 #350
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[최단 경로] 2171039 이채원 #350
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코드리뷰완료 |
Dong-droid
reviewed
May 30, 2023
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+50
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| //정점 별로 모든 정점에 대한 최단경로를 구함. | ||
| for (int i = 1; i <= n; i++) { | ||
| dist[i] = dijkstra(i, n, graph); | ||
| } | ||
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| vector<int> dist_sum = dijkstra(x, n, graph); //파티장에서 모든 정점으로 돌아오는 최단 경로 | ||
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| //파티장에서 돌아오는 정점에 정점에서 파티장으로 가는 최단경로의 합을 구함. | ||
| for (int i = 1; i <= n; i++) { | ||
| dist_sum[i] += dist[i][x]; | ||
| max_dist = max(max_dist, dist_sum[i]); //최댓값 갱신 | ||
| } | ||
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Dong-droid
reviewed
May 30, 2023
Comment on lines
+24
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+36
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| for (int i = 1; i <= n; i++) { | ||
| if (graph[node][i] == INF) { //연결되어 있지 않는 노드일 때 | ||
| q.push(i); | ||
| } | ||
| } | ||
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| while (!q.empty()) { | ||
| if (graph[q.front()][node] == INF) { //반대로도 연결되어 있지 않을 때 -> 키 순서 파악 불가. | ||
| return false; | ||
| } | ||
| q.pop(); | ||
| } |
There was a problem hiding this comment.
P3. queue를 사용해도 되지만
for(int i = 1; i <=n; i++)
if(graph[node][i] == INF && graph[i][node] == INF) return false;
를 사용해서 for문으로도 구현할 수 있습니다~
jk0527
reviewed
Jun 2, 2023
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인적사항
학번: 2171039
이름: 이채원
과제 제출
기존 제출 : 1238, 2458
추가 제출 : 15685