Description
Building and maintaining roads among communities in the far North is an expensive business. With this in mind, the roads are build such 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.
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
Input to the problem 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.
Output
You are to output 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
22
给定一颗棵树,求最长路径
思路：随机找一个节点u,DFS求出u的最远点m，然后再DFS求出m的最远点n，之后m-n就是最长路径
2.还可以用DP写，d(i)表示以i为根节点的最大路径值，=max(d(j)+1),j为i的子节点，取出最大和次大的d(j) +2；
1 #include <cstring> 2 #include <iostream> 3 #include <algorithm> 4 #include <cstdio> 5 #include <vector> 6 #define Max 10005 7 using namespace std; 8 vector <int> tree[Max],len[Max]; 9 int d[Max],vis[Max]; 10 bool flag=0; 11 int an; 12 int dfs(int node,int &ans) 13 { 14 int i,j; 15 int maxn=0; 16 int t,index; 17 vis[node]=1; 18 for(i=0;i<tree[node].size();i++) 19 { 20 if(vis[tree[node][i]]) 21 continue; 22 dfs(tree[node][i],t); 23 if((t+len[node][i])>maxn) 24 { 25 // cout<<t<<endl; 26 maxn=t+len[node][i]; 27 index=i; 28 } 29 } 30 ans=maxn; 31 return i; 32 } 33 void dfs1(int node,int sum) 34 { 35 int i; 36 vis[node]=1; 37 if(flag) 38 return; 39 if(sum==0) 40 { 41 flag=1; 42 an=node; 43 return; 44 } 45 for(i=0;i<tree[node].size();i++) 46 { 47 if(sum>=len[node][i]&&vis[tree[node][i]]==0) 48 dfs1(tree[node][i],sum-len[node][i]); 49 } 50 } 51 int main() 52 { 53 int i,j; 54 int a,b,val,p=0; 55 freopen("in.txt","r",stdin); 56 bool flag=0; 57 for(i=0;i<Max;i++) 58 tree[i].clear(),len[i].clear(); 59 while(scanf("%d%d%d",&a,&b,&val)!=EOF) 60 { 61 tree[a].push_back(b); 62 tree[b].push_back(a); 63 len[a].push_back(val); 64 len[b].push_back(val); 65 } 66 memset(vis,0,sizeof(vis)); 67 dfs(1,p); 68 memset(vis,0,sizeof(vis)); 69 dfs1(1,p); 70 memset(vis,0,sizeof(vis)); 71 dfs(an,p); 72 cout<<p<<endl; 73 }