【HDU 3848 CC On The Tree】+ 递归

CC On The Tree
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 125536/65536 K (Java/Others)
Total Submission(s): 1949 Accepted Submission(s): 681

Problem Description
CC lives on the tree which has N nodes.On every leaf of the tree there is an apple(leaf means there is only one branch connect this node ) .Now CC wants to get two apple ,CC can choose any node to start and the speed of CC is one meter per second.
now she wants to know the shortest time to get two apples;

Input
Thers are many cases;
The first line of every case there is a number N(2<=N<=10000)
if n is 0 means the end of input.
Next there is n-1 lines,on the i+1 line there is three number ai,bi,ci
which means there is a branch connect node ai and node bi.
(1<=ai, bi<=N , 1<=ci<=2000)
ci means the len of the branch is ci meters ;

Output
print the mininum time to get only two apple;

Sample Input
7
1 2 1
2 3 2
3 4 1
4 5 1
3 6 3
6 7 4
0

Sample Output
5

两个叶子节点必有一个相同的父节点，更新父节点的最短和次短叶子节点

AC代码：

#include<cstdio>#include<algorithm>using namespace std;const int K = 1e4 + 10;const int inf = 0x3f3f3f3f / 2;struct node{    int to,vl,next;}st[K * 2];int num,n,pr[K],f[K],s[K],head[K];void add(int a,int b,int c){    st[num].to = b,st[num].vl = c,st[num].next = head[a],head[a] = num++;}void dfs(int a,int b){    if(pr[a] == 1){ // 叶子节点        f[a] = 0;        return ;    }    for(int i = head[a]; i != -1; i = st[i].next){        int ma = st[i].to;        if(ma == b) continue;        dfs(ma,a);        int mb = st[i].vl;        if(f[a] - f[ma] > mb) s[a] = f[a],f[a] = f[ma] + mb; // 最短        else if(s[a] - f[ma] > mb) s[a] = f[ma] + mb; // 次短    }}int main(){    int a,b,c;    while(~scanf("%d",&n),n){        num = 0;        fill(head,head + n + 1,-1);        fill(pr,pr + n + 1,0);        fill(f,f + n + 1,inf);        fill(s,s + n + 1,inf);        for(int i = 1; i < n; i++){            scanf("%d %d %d",&a,&b,&c);            add(a,b,c),add(b,a,c);            pr[a]++,pr[b]++;        }        if(n == 2){            printf("%d\n",c);            continue;        }        for(int i = 1; i <= n; i++)            if(pr[i] > 1){                dfs(i,-1);                break;            }        int ans = inf;        for(int i = 1; i <= n; i++)            if(s[i] && f[i])                ans = min(ans,s[i] + f[i]);        printf("%d\n",ans);    }    return 0;}