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.
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
Analysis
Solution 1:
树的结点往上长型dp
Note:
- 图的邻接表模板
- 字符串的处理
Solution 2:
两次DFS
Suppose a and b are the endpoints of the path in the tree which achieve the diameter.
Let s be any vertex in T,then a or b is the vertex whose distance from s is the greatest.
Code
#include<bits/stdc++.h>
using namespace std;
int dp1[10001],dp2[10001],head[10001];
int cnt=0,maxi=0;
struct edge{int v,w,next;}Edge[20001];
inline void add_edge(int u,int v,int w){
Edge[++cnt]={v,w,head[u]};head[u]=cnt;
}
void dfs(int s,int pre){
for(int i=head[s];i;i=Edge[i].next){
int v=Edge[i].v;
if(v==pre)continue;dfs(v,s);
if(dp1[v]+Edge[i].w>dp1[s])dp2[s]=dp1[s],dp1[s]=dp1[v]+Edge[i].w;
else if(dp1[v]+Edge[i].w>dp2[s])dp2[s]=dp1[v]+Edge[i].w;
}maxi=max(maxi,dp1[s]+dp2[s]);
}
int main(){
int a,b,c,flag;char m[20];
while(1){
memset(head,-1,sizeof(head));memset(dp1,0,sizeof(dp1));
flag=cnt=maxi=0;memset(dp2,0,sizeof(dp2));
while(gets(m)!=NULL&&m[0]!=' '){
sscanf(m,"%d%d%d",&a,&b,&c);
add_edge(a,b,c),add_edge(b,a,c);flag=1;
}
if(flag==1){
dfs(1,0);
printf("%d
",maxi);
}if(m[0]!=' ')break;
}
return 0;
}