HDU 2196 Computer( 樹上節點的最遠距離 )

標籤：

Computer

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4080    Accepted Submission(s): 2043

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 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. 

 

InputInput 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. 

 

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  經典的問題，第一次Dp先處理好子樹的 最遠 ， 次遠距離（ 有一邊[ u , v , w ] , u 的次遠可能是最遠路上兒子 v 的最遠（ u的次遠 + w ） )~~解決的時候考慮 temp 是來自父親路徑上的最長路，然後對 兒子分成 是否是最遠路上的~ 來求解答案~ 

#include <bits/stdc++.h>using namespace std ;const int N = 10010 ;int dp[N][2] , son[N][2] , ans[N] , n ;int eh[N] , et[N<<1] , nxt[N<<1] , ew[N<<1] , tot ;void init() {    memset( eh , -1 , sizeof eh );    tot = 0 ;} void addedge( int u , int v , int w ) {    et[tot] = v ; ew[tot] = w ; nxt[tot] = eh[u] ; eh[u] = tot++ ;    et[tot] = u ; ew[tot] = w ; nxt[tot] = eh[v] ; eh[v] = tot++ ;}int Dp( int u , int fa ) {    for( int i = eh[u] ; ~i ; i = nxt[i] ) {        int v = et[i] , w = ew[i] ;        if( v == fa ) continue ;        int tmp = Dp( v , u ) + w ;        if( tmp > dp[u][1] ) {            dp[u][1] = tmp ;            son[u][1] = v ;        }        if( dp[u][1] > dp[u][0] ) {            swap( dp[u][1] , dp[u][0] ) ;            swap( son[u][1] , son[u][0]) ;        }    }    return dp[u][0] ;}void Solve( int u , int fa , int tmp ) {    ans[u] = max( dp[u][0] , tmp ) ;    for( int i = eh[u] ; ~i ; i = nxt[i] ) {        int v = et[i] , w = ew[i] ;        if( v == fa ) continue ;        if( v == son[u][0] ) {            Solve( v , u , max( dp[u][1] , tmp ) + w ) ;        } else {            Solve( v , u , max( dp[u][0] , tmp ) + w ) ;        }    }}int main () {    while( ~scanf("%d",&n) ) {        init();        for( int i = 2 ; i <= n ; ++i ) {            int v , w ; scanf("%d%d",&v,&w);            addedge( i , v , w ) ;         }        memset( dp , 0 , sizeof dp ) ;        Dp( 1 , 0 );        //for( int i = 1 ; i <= n ; ++i ) cout << i << ‘ ‘ << dp[i][0] << ‘ ‘ << dp[i][1] << endl ;        Solve( 1 , 0 , 0 );        for( int i = 1 ; i <= n ; ++i ) printf("%d\n",ans[i]);    }    return 0 ;}

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HDU 2196 Computer( 樹上節點的最遠距離 )