D - Tree and Hamilton Path
Time limit : 2sec / Memory limit : 256MB
Score : 1100 points
Problem Statement
There is a tree with N vertices, numbered 1 through N. The i-th edge in this tree connects Vertices Ai and Bi and has a length of Ci.
Joisino created a complete graph with N vertices. The length of the edge connecting Vertices u and v in this graph, is equal to the shortest distance between Vertices u and v in the tree above.
Joisino would like to know the length of the longest Hamiltonian path (see Notes) in this complete graph. Find the length of that path.
Notes
A Hamiltonian path in a graph is a path in the graph that visits each vertex exactly once.
Constraints
- 2≤N≤105
- 1≤Ai<Bi≤N
- The given graph is a tree.
- 1≤Ci≤108
- All input values are integers.
Input
Input is given from Standard Input in the following format:
N A1 B1 C1 A2 B2 C2 : AN−1 BN−1 CN−1
Output
Print the length of the longest Hamiltonian path in the complete graph created by Joisino.
Sample Input 1
5 1 2 5 3 4 7 2 3 3 2 5 2
Sample Output 1
38
The length of the Hamiltonian path 5 → 3 → 1 → 4 → 2 is 5+8+15+10=38. Since there is no Hamiltonian path with length 39 or greater in the graph, the answer is 38.
Sample Input 2
8 2 8 8 1 5 1 4 8 2 2 5 4 3 8 6 6 8 9 2 7 12
Sample Output 2
132
依然是按照官方题解的做法写的。DFS的方法比较巧妙，c数组 按照DFS的“方向”，记录某一节点在该方向上（包括自身）所有后代节点个数。这样统计官方题解中的si就非常方便了。
1 #include <iostream> 2 #include <string> 3 #include <algorithm> 4 #include <cstring> 5 #include <cstdio> 6 #include <cmath> 7 #include <queue> 8 #include <set> 9 #include <map> 10 #include <list> 11 #include <vector> 12 #include <stack> 13 #define mp make_pair 14 #define MIN(a,b) (a>b?b:a) 15 //#define MAX(a,b) (a>b?a:b) 16 typedef long long ll; 17 typedef unsigned long long ull; 18 const int MAX=1e5+5; 19 const int INF=1e9+5; 20 using namespace std; 21 //const int MOD=1e9+7; 22 typedef pair<ll,int> pii; 23 typedef pair<ll,ll> pll; 24 const double eps=0.00000001; 25 ll n,an,m=INF; 26 vector<pll>edge[MAX]; 27 ll c[MAX]; 28 void calc(ll now,ll pre) 29 { 30 c[now]=1; 31 for(auto &X:edge[now]) 32 { 33 ll to,val; 34 tie(to,val)=X; 35 if(to==pre) 36 continue; 37 calc(to,now); 38 c[now]+=c[to]; 39 } 40 } 41 void solve(ll now,ll pre,ll len)//从pre连到now的一条长为len的边 42 { 43 an+=2*len*min(c[now],n-c[now]); 44 pll mx={-1,-1};//now的最大子树的节点数 及连向其的边的权值 45 for(auto &X:edge[now]) 46 { 47 ll to,val; 48 tie(to,val)=X; 49 if(to==pre) 50 continue; 51 mx=max(mx,{c[to],val}); 52 } 53 mx=max(mx,{n-c[now],len}); 54 if(2*mx.first==n) 55 m=min(m,mx.second); 56 else if(2*mx.first<n) 57 { 58 for(auto &X:edge[now]) 59 m=min(m,X.second); 60 } 61 for(auto &X:edge[now]) 62 { 63 ll to,val; 64 tie(to,val)=X; 65 if(to==pre) 66 continue; 67 solve(to,now,val); 68 } 69 } 70 71 int main() 72 { 73 scanf("%lld",&n); 74 for(ll i=1;i<n;i++) 75 { 76 ll A,B,C; 77 scanf("%lld%lld%lld",&A,&B,&C); 78 edge[A].push_back({B,C}); 79 edge[B].push_back({A,C}); 80 } 81 calc(1,0); 82 solve(1,0,0); 83 printf("%lld ",an-m); 84 85 }