HDU3848--CC On The Tree【BFS】

CC On The Tree

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 125536/65536 K (Java/Others)
Total Submission(s): 1726    Accepted Submission(s): 617

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

 

注意：建图的时候注意标记每个节点的入度，入度为1的肯定为叶子节点，从一个叶子节点开始 BFS ，知道找到下一个叶子节点结束。

#include <cstdio>
#include <cstring>
#include <iostream>
#include <queue>
#include <cstdlib>
#define min(a , b) (a > b ? b : a)
#define INF 0x3f3f3f3f
using namespace std;
int n, num;
struct node {
    int u;
    int v;
    int w;
    int next;
};

int head[11000], vis[11000], cnt, du[11000]; 
node edge[22000];

struct mod{
    int x, step;
    friend bool operator < (mod s1, mod s2)
    {
        return s1.step > s2.step;
    }
};

void add(int u, int v, int w){
    edge[cnt].u = u;
    edge[cnt].v = v;
    edge[cnt].w = w;
    edge[cnt].next = head[u];
    head[u] = cnt++;
}

bool judge(mod s){
    if(vis[s.x]) return 0;
    if(s.step > num) return 0;
        return 1;
}

mod s1,temp;

int bfs(int x){
    s1.x = x;
    s1.step = 0;
    priority_queue<mod>q;
    q.push(s1);
    memset(vis, 0 ,sizeof(vis));
    vis[s1.x] = 1;
    while(!q.empty()){
        s1 = q.top();
        q.pop();
        int u = s1.x;
        for(int i = head[u]; i != -1; i = edge[i].next){
            int u = edge[i].u;
            int v = edge[i].v;
            int w = edge[i].w;
            temp.x = v;
            temp.step = s1.step + w;
            if(!judge(temp)) continue;
            if(du[v] == 1)
                return temp.step;
            vis[temp.x] = 1;
            q.push(temp);
        }
    }
    return num;
}

int main (){
    while(scanf("%d", &n),n){
        memset(du, 0 ,sizeof(du));
        memset(head, -1, sizeof(head));
        cnt  = 0;
        for(int i = 0; i < n - 1; ++i){
            int u, v, w;
            scanf("%d%d%d", &u, &v, &w);
            add(u, v, w);
            add(v, u, w);
            du[u]++;
            du[v]++;
        }
        num = INF;
            for(int i = 1 ; i <= n; ++i){
                if(du[i] == 1){
                    int sum = bfs(i);
                    num = min(num, sum);
                }
            }
            printf("%d\n", num);
    }
    return 0;
}