Roads in the North - UVa 10308 树形dp

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在一片连接着多个村庄的树形地图上，任务是找出相距最远的两个村庄。通过遍历每座村庄的子树并找到到叶节点距离最大的两点，最终计算出这两点之间的道路距离。

Problem I

Roads in the North

Input: standard input

Output: standard output

Time Limit: 2 seconds

Memory Limit: 32 MB

Building and maintaining roads among communities in the far North is an expensive business. With this in mind, the roads are built in such a way that there is only one route from a village to a village that does not pass through some other village twice.

Given is an area in the far North comprising a number of villages and roads among them such that any village can be reached by road from any other village. Your job is to find the road distance between the two most remote villages in the area.

The area has up to 10,000 villages connected by road segments. The villages are numbered from 1.

Input

The input contains several sets of input. Each set of input is a sequence of lines, each containing three positive integers: the number of a village, the number of a different village, and the length of the road segment connecting the villages in kilometers. All road segments are two-way. Two consecutive sets are separated by a blank line.

Output

For each set of input, you are to output a single line containing a single integer: the road distance between the two most remote villages in the area.

Sample Input

5 1 6

1 4 5

6 3 9

2 6 8

6 1 7

Sample Output

题意：在一棵树上找距离最远的两个点。

思路：每个点找到它的子树中到叶节点距离最大的两个，答案就可能为这两个之和。

AC代码如下：

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
struct node
{
    int v,w,next;
}edge[20010];
int Head[10010],num,dp[10010][2],maxn;
void add(int u,int v,int w)
{
    edge[++num].v=v;
    edge[num].w=w;
    edge[num].next=Head[u];
    Head[u]=num;
}
char s[110];
void dfs(int f,int u)
{
    int j,v,w;
    for(j=Head[u];j!=0;j=edge[j].next)
       if(edge[j].v!=f)
       {
           v=edge[j].v;
           dfs(u,v);
           w=edge[j].w+dp[v][0];
           if(w>dp[u][0])
           {
               dp[u][1]=dp[u][0];
               dp[u][0]=w;
           }
           else if(w>dp[u][1])
             dp[u][1]=w;
       }
    maxn=max(maxn,dp[u][0]+dp[u][1]);
}
int main()
{
    int i,j,k,u,v,w;
    while(gets(s))
    {
        if(s[0]=='\0')
        {
            dfs(0,1);
            printf("%d\n",maxn);
            memset(Head,0,sizeof(Head));
            memset(dp,0,sizeof(dp));
            maxn=0;num=0;
            continue;
        }
        sscanf(s,"%d%d%d",&u,&v,&w);
        add(u,v,w);
        add(v,u,w);
    }
    dfs(0,1);
    printf("%d\n",maxn);
}