HDU 3534

Tree

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1288    Accepted Submission(s): 399

Problem Description

 

　　In the Data structure class of HEU, the teacher asks one problem: How to find the longest path of one tree and the number of such longest path?

 

Input

 

　　There are several test cases. The first line of each case contains only one integer N, means there are N nodes in the tree. N-1 lines follow, each line has three integers w,v and len, indicate that there is one edge between node w and v., and the length of the edge is len.

Output

 

　　For each test case, output the length of longest path and its number in one line.

 

Sample Input

 

4

1 2 100

2 3 50

2 4 50

4

1 2 100

2 3 50

3 4 50

 

Sample Output

 

150 2

200 1

 

通过这几天的看题目，做题目，发现树形DP很有规律，几乎知道思路，就可以写了

 

题意：

　　给你一棵树，求最长路径，并且求出有多少条这样的路径

分析：

　　任意两点只有一条路径最短路径，ok，所以可以用数组来标记当前状态下最长路径是多少，有多少条

 

AC代码：

# include <bits/stdc++.h>
using namespace std;
const int MAX = 50001;
struct Node
{
    int to;
    int next;
    int len;
}tree[MAX * 2];
int head[MAX], Len[MAX], node[MAX];
int tol = 0;
int Max = 0, jishu = 0;

void add(int a, int b, int len)
{
    tree[tol].to = b;
    tree[tol].next = head[a];
    tree[tol].len = len;
    head[a] = tol++;
}
void dfs(int root, int f)
{
    Len[root] = 0;
    node[root] = 1;
    for(int i = head[root]; i != -1; i = tree[i].next)
    {
        int son = tree[i].to;
        if(son == f)
            continue;
        dfs(son, root);
        
        int tep = Len[son] + tree[i].len;  
        if(tep + Len[root] > Max)//最长边经过son
        {  
            jishu = node[son] * node[root];  
            Max = tep + Len[root];  
        }  
        else if(tep + Len[root] == Max)  
            jishu += node[son] * node[root];

        if(Len[root] < tep)  
        {
            Len[root] = tep;  
            node[root] = node[son];  
        }  
        else if(Len[root] == tep)  
            node[root] += node[son];      
    }    
}
int main()
{
    int n;
    while(scanf("%d", &n) != EOF)
    {
        tol = 0;
        memset(head, -1, sizeof(head));
        
        int a, b, len;
        for(int i = 1; i < n; i++)
        {
            scanf("%d%d%d", &a, &b, &len);
            add(a, b, len);
            add(b, a, len);
        }
        Max = 0, jishu = 0;
        dfs(1, -1);
        printf("%d %d
", Max, jishu);
    }
    return 0;
}