Nuclear P1099 Tree Network - Analog + tree

Nuclear P1099 tree network

Unrooted tree, to find a path length of not more than s in diameter, so that the farthest point from this shortest path;

First, find the diameter dfs twice (the second time when we must find the right father [] is cleared)

Found enumerated in diameter does not exceed the length s of the chain, ans ends of the lower bound to the minimum diameter point distance of this chain;

Then the points on the diameter are marked, ask again what else to point distance diameter.

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<algorithm>
 4 using namespace std;
 5 const int maxn=620;
 6 int n,s;
 7 int pre[maxn],last[maxn],other[maxn],len[maxn],l;
 8 
 9 void add(int x,int y,int z)
10 {
11     l++;
12     pre[l]=last[x];
13     last[x]=l;
14     other[l]=y;
15     len[l]=z;
16 }
17 
18 int dis[maxn],father[maxn],vis[maxn];
19 int l1,l2;
20 
21 void dfs(int x)
22 {
23     for(int p=last[x];p;p=pre[p])
24     {
25         int v=other[p];
26         if(v==father[x]||vis[v]) continue;
27         father[v]=x;
28         dis[v]=dis[x]+len[p];
29         dfs(v);
30     }
31 }
32 
33 
34 int main()
35 {
36     scanf("%d%d",&n,&s);
37     for(int i=1;i<n;i++)
38     {
39         int x,y,z;
40         scanf("%d%d%d",&x,&y,&z);
41         add(x,y,z);
42         add(y,x,z);
43     }    
44     
45     dis[1]=0; dfs(1);    
46     l1=1;
47     for(int i=1;i<=n;i++) if(dis[i]>dis[l1]) l1=i;
48     memset(father,0,sizeof(father));
49     dis[l1]=0; dfs(l1); l2=1;
50     for(int i=1;i<=n;i++) if(dis[i]>dis[l2]) l2=i;
51     
52     //printf("??%d %d\n",l1,l2);
53     
54     int ans=2147483647,j=l2;
55     for(int i=l2;i;i=father[i])
56     {
57         while(father[j]&&dis[i]-dis[father[j]]<=s) j=father[j];
58         ans=min(ans,max(dis[l2]-dis[i],dis[j]));
59     }
60     for(int i=l2;i;i=father[i]) vis[i]=1;
61     for(int i=l2;i;i=father[i]) 
62     {
63         dis[i]=0;
64         dfs(i);
65     }
66     for(int i=1;i<=n;i++) ans=max(ans,dis[i]);
67     printf("%d",ans);
68     return 0;
69 }

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