Computer
Time limit:1000/1000 MS (java/others) Memory limit:32768/32768 K (java/others)
Total submission (s): 4440 Accepted Submission (s): 2236
Problem Descriptiona School bought the first computer some time ago (so this computer ' s ID is 1). During The recent years the school bought N-1 new computers. Each new computer is connected to one of settled earlier. Managers of school is anxious about slow functioning of the net and want to know the maximum distance Si for which i-th C Omputer needs to send signal (i.e. length of cable-the most distant computer). You need to provide this information.
hint:the example input is corresponding to this graph. And from the graph, you can see that the computer 4 are farthest one from 1, so S1 = 3. Computer 4 and 5 is the farthest ones from 2, so S2 = 2. Computer 5 is the farthest one from 3, so S3 = 3. We also get S4 = 4, S5 = 4.
Inputinput file contains multiple test cases. In each case there was natural number N (n<=10000) in the first line, followed by (N-1) lines with descriptions of Compu Ters. I-th line contains-natural numbers-number of computer, to which i-th computer is connected and length of cable used for connection. Total length of cable does not exceed 10^9. Numbers in lines of input is separated by a space.
Outputfor each case output N lines. I-th line must contain number Si for i-th computer (1<=i<=n).
Sample Input51 12 13 11 1
Sample Output32344 has just begun to use violence, on every node to BFS a predictable LTE test instructions: Buy a computer, every new computer to connect it to the purchase of the last computer ask all the nodes as the starting point of the longest path, The given test data means that: Two number A, b means that a computer is connected to the I+1 computer, the required cable length is B (I refers to line i) 2 in the test data 1 means that 2 and 3 are connected with a weighted value of 1. 3 1 means that computer 3 is connected to the computer 4 with a weight of 1; Solving the problem: using three BFs to find the longest road two endpoints U and v because each new computer is connected to the previous computer, each computer's connection is not forked, That is, the longest road of the node is either the distance to the endpoint u or the distance to the endpoint V (that is, take the larger one), first two times BFS to find an endpoint record the distance from each point to this endpoint, and then use the BFS to find another endpoint, Then record the distance from each point to this endpoint finally traverse each point to find the longest
#include <stdio.h> #include <string.h> #include <queue> #define MAX 40100#define maxn (x, y) (x>y?x:y ) using namespace Std;int head[max];int vis[max],dis[max];int n,m,ans,ant;int sum,beg,en;int a[max],b[max];struct node{ int U,v,w;int Next; edge[max];void Add (int u,int v,int W) {Edge[ans].u=u;edge[ans].v=v;edge[ans].w=w;edge[ans].next=head[u];head[u]=ans ++;} void Getmap () {int I,j,a,b;ans=0;memset (head,-1,sizeof (head)), for (i=2;i<=n;i++) {scanf ("%d%d", &a,&b); Add (a,i,b); add (i,a,b);}} void BFs (int sx) {int I,j;queue<int>q;sum=0;memset (vis,0,sizeof (Vis)); memset (dis,0,sizeof (dis)); vis[sx]=1; Beg=sx;q.push (SX), while (!q.empty ()) {int Top=q.front (), Q.pop (), for (i=head[top];i!=-1;i=edge[i].next) {int k=edge[i ].v;if (!vis[k]) {Vis[k]=1;dis[k]=dis[top]+edge[i].w;q.push (k);} if (Sum<dis[k]) {sum=dis[k];beg=k;}}}} void Solve () {int I,j;memset (a,0,sizeof (a)), memset (b,0,sizeof (b)), BFS (1), BFS (Beg), en=beg;//find first endpoint for (i=1;i<=n i++) a[i]=dis[i];//record the distance from each point to this endpoint bfs (en);//Find another endpoint for (i=1;i<=n;i++) b[i]=dis[i];//records the distance of each point to this endpoint for (i=1;i<=n;i++) {ant=0; ANT=MAXN (A[i],b[i]); printf ("%d\n", Ant);}} int main () {while (scanf ("%d", &n)!=eof) {getmap (); Solve ();} return 0;}
Hdoj 2196 Computer "the diameter of the tree asks for all the longest path starting from any node"
