HDU 2196 computer (tree DP)

Computer Time Limit: 1000/1000 MS (Java/others) memory limit: 32768/32768 K (Java/Others)
Total submission (s): 3075 accepted submission (s): 1561

Problem description

A 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 was connected to one of settled earlier. managers of school are anxious about slow functioning of the net and want to know the maximum distance Si for which I-th computer needs to send signal (I. e. length of cable to 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 is farthest one from 1, so S1 = 3. computer 4 and 5 are 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.

Input

Input file contains multiple test cases. in each case there is natural number n (n <= 10000) In the first line, followed by (N-1) lines with descriptions of computers. i-th line contains two 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 are separated by a space.

Output

For each case output n lines. I-th line must contain number si for I-th computer (1 <= I <= N ).

Sample Input

51 12 13 11 1

Sample output

32344

: All computers are connected to a tree structure and each node must be located at the maximum distance from the node.

If each node is the root node and a DFS is performed, it is easy to find the longest distance. However, if it is only violent, time is not enough,

However, we can use the DP idea to avoid repeated computing sub-problems. Using this pruning method, the efficiency is quite high.

My implementation method: Use each node as the root node for a DFS, and use DP to save the longest distance that can be reached through a certain edge.

Edge storage is bidirectional. Each edge is assigned a sequence number. // DP [x] below, and X is the edge number

For example, first select any node and select 1 as the root node for DFS. Then we can find the side passing through the side 2 Side 1... side 4 .... The maximum distance that can be reached: DP [1], d [2]… DP [4] ..

2 is the root node DFS, because Edge 4 can reach the longest distance DP [4] To Know .. DP [7], DP [2] also know that the distance from node 2 is equal to max (DP [4], DP [7], DP [2] + dis [3]), // dis [3] indicates the length of the third edge

view code#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int N = 10010;int n, dp[N<<1], pre[N];struct edge{    int u, v, w, p;    edge() {}    edge(int u, int v, int w, int p):u(u), v(v), w(w), p(p) {}}e[N<<1];int ecnt = 0;int dfs(int u, int p){    int ans = 0;    for(int i=pre[u]; ~i; i=e[i].p)    {        int v = e[i].v;        if(v==p) continue;        if(!dp[i]) dp[i] = dfs(v, u)+e[i].w;        ans = max(ans , dp[i]);    }    return ans;}int main(){//    freopen("in", "r", stdin);    while(scanf("%d", &n)>0)    {        memset(dp, 0, sizeof(dp));        memset(pre, -1, sizeof(pre));        int v, w;        ecnt = 0;        for(int i=2; i<=n; i++)        {            scanf("%d%d", &v, &w);            e[ecnt] = edge(i, v, w, pre[i]);            pre[i] = ecnt++;            e[ecnt] = edge(v, i, w, pre[v]);            pre[v] = ecnt++;        }        for(int i=1; i<=n; i++)            printf("%d\n", dfs(i,-1));    }    return 0;}