{intto;intval;intnext; }; Edge edge[20005];inthead[10005];voidadd(intu,intv,intval) { edge[cnt].to=v; edge[cnt].val=val; edge[cnt].next=head[u]; head[u]=cnt++; }intdfs(intx,intf,intval) {intsum=1;///以x为根的子树个数,自己算1个for(inti=head[x];i!=-1;i=edge[i]...
(startx, starty, 0); cout << "shortest path = " << min << endl; system("pause"); return 0; } void dfs(int x, int y, int step) { int i, tx, ty; int next[4][2] = { {0,1},{1,0},{0,-1},{-1,0} };//设置访问方向优先级 if (x == p && y == q) {//...
牛客Shortest Path (dfs+思维) 非常巧妙的转换。首先我们可以观察性质,一条边不可能出现两次,因为显然可以被更好的方案替代 之后,每个边就存在选或者不选两种情况 如果以子树的节点为偶数,那么就不需要,否则需要,做一下dfs即可 View Code
for next in graph[vertex]: if next == end: yield path + [next] else: queue.append((next, path+[next])) def shortest_path_bfs(graph, start, end): try: return next(bfs_paths(graph, start, end)) except StopIteration: return None def main(): print(shortest_path_bfs(graph, 'A', ...
{shortestDistance=Math.min(shortestDistance,distance);return;}// 遍历邻接节点for(intneighbor:graph.getAdjacencyList().getOrDefault(node,newArrayList<>())){dfs(graph,neighbor,distance+1);// 当前距离加一}// 结束访问,回溯visited.remove(node);}// 获取最短距离publicintgetShortestDistance(){return...
col]==1) return ; //输入错误 if(startP==endP){ //起点即终点 shortestPath.push_back(startP); return; } //mark标记每一个节点的前驱节点,如果没有则为(-1,-1),如果有,则表示已经被访问 Point** mark=new Point*[m]; for(int i=0;i<m;++i){ mark[i]=new Point[n]; } queue<Point...
dfs(startNode, endNode, visited, 0, shortestPath, &minLength); 打印最短路径 printf("Shortest path length: %d\n", minLength); printf("Shortest path: "); for (int i = 0; i <= endNode; i++) { printf("%d ", shortestPath[i]); } free(visited); free(shortestPath); return 0; ...
public booleanareSentencesSimilarTwo(String[]words1,String[]words2,String[][]pairs){if(words1.length!=words2.length){returnfalse;}// construct a map by pairs using dfs// key word, value the word's neighbours, is an undirected graphMap<String,Set<String>>map=newHashMap<>();for(String[...
for(inti=0;i<n;i++) { a[i]=in.nextInt(); } intans=process(a,0,n-1); pw.println(ans); pw.flush(); } } publicstaticintprocess(int[]arr,intl,intr){ if(l==r){ return0; } intmid=l+((r-l)>>1); returnprocess(arr,l,mid)+process(arr,mid+1,r)+merge(arr,l,mid,r...
// iterate the nodes which are already in the queueintsize=queue.size();for(inti=0;i<size;++i){Node cur=the first node in queue;returnstepifcur is target;for(Node next:the neighbors of cur){add next to queue;}remove the first node from queue;}}return-1;// there is no path ...