The diameter of the POJ 1849 tree or the tree dp

Two

Time Limit:1000 MS		Memory Limit:30000 K
Total Submissions:1081		Accepted:517

Description

The city consists of intersections and streets that connect them.

Heavy snow covered the city so the mayor Milan gave to the winter-service a list of streets that have to be cleaned of snow. these streets are chosen such that the number of streets is as small as possible but still every two intersections to be connected I. e. between every two intersections there will be exactly one path. the winter service consists of two snow plovers and two drivers, Mirko and Slavko, and their starting position is on one of the intersections.

The snow plover burns one liter of fuel per meter (even if it is driving through a street that has already been cleared of snow) and it has to clean all streets from the list in such order so the total fuel spent is minimal. when all the streets are cleared of snow, the snow plovers are parked on the last intersection they visited. mirko and Slavko don't have to finish their plowing on the same intersection.

Write a program that calculates the total amount of fuel that the snow plovers will spend.

Input

The first line of the input contains two integers: N and S, 1 <= N <= 100000, 1 <= S <= N. N is the total number of intersections; S is ordinal number of the snow plovers starting intersection. intersections are marked with numbers 1... n.

Each of the next N-1 lines contains three integers: A, B and C, meaning that intersections A and B are directly connected by a street and that street's length is C meters, 1 <= C <= 1000.

Output

Write to the output the minimal amount of fuel needed to clean all streets.

Sample Input

5 21 2 12 3 23 4 24 5 1

Sample Output

6

There are n regions, n-1-1 sides, and two vehicles in a certain place. They want to start from the starting point and clean all the streets, that is, the edge of the tree, and find the minimum path.

At first glance, I have no idea. I can only think about the diameter of the tree. I can directly take the double-diameter of all edges as well as ac. However, I don't understand the principle and cannot prove the correctness.

Code:

/*************************************** * ******** Author: xianxingwuguanCreated Time: 2014-2-6 4: 48: 24 File Name: 3. cpp *************************************** * ********/# pragma comment (linker, "/STACK: 102400000,102400000") # include
 
  
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             Using namespace std; # define INF 0x3f3f3f # define eps 1e-8 # define pi acos (-1.0) typedef long ll; const int maxn = 100100; int head [maxn], dis [maxn], tol, n; struct node {int next, to, val; node () {}; node (int _ next, int _ to, int _ val ): next (_ next), to (_ to), val (_ val) {}} edge [3 * maxn]; void add (int u, int v, int val) {edge [tol] = node (head [u], v, val); head [u] = tol ++;} int bfs (int & s) {memset (dis, -1, sizeof (dis); dis [s] = 0; queue
              
                Q; q. push (s); while (! Q. empty () {int u = q. front (); q. pop (); for (int I = head [u]; I! =-1; I = edge [I]. next) {int v = edge [I]. to; if (dis [v]! =-1) continue; dis [v] = dis [u] + edge [I]. val; q. push (v) ;}} int mx = 0; for (int I = 1; I <= n; I ++) if (dis [I]> mx) mx = dis [I], s = I; return mx;} int main () {// freopen ("data. in "," r ", stdin); // freopen (" data. out "," w ", stdout); int I, j, k, m, p; while (~ Scanf ("% d", & n, & m) {memset (head,-1, sizeof (head); tol = 0; int sum = 0; for (k = 1; k
               
                

                
The following is a tree-like dp. dp [u] [I] indicates that the subtree rooted in u has the cost of an I car, and then the process of u and the child node backpack and the new dp value.
                
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/* ***********************************************Author :xianxingwuguanCreated Time :2014-2-6 5:53:24File Name :3.cpp************************************************ */#pragma comment(linker, "/STACK:102400000,102400000")#include 
                 
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                             using namespace std;#define INF 0x3f3f3f3f#define eps 1e-8#define pi acos(-1.0)typedef long long ll;const int maxn=100100;int head[maxn],dp[maxn][3],tol;struct node{int next,to,val;node(){};node(int _next,int _to,int _val):next(_next),to(_to),val(_val){}}edge[3*maxn];void add(int u,int v,int val){edge[tol]=node(head[u],v,val);head[u]=tol++;}void dfs(int u,int fa){for(int i=head[u];i!=-1;i=edge[i].next){int v=edge[i].to;if(v==fa)continue;dfs(v,u);int ret[3];for(int j=0;j<3;j++){ ret[j]=dp[u][j]; dp[u][j]=INF;}int tmp[3]={2*edge[i].val,edge[i].val,2*edge[i].val};for(int j=2;j>=0;j--)for(int k=0;k<=j;k++)dp[u][j]=min(dp[u][j],ret[j-k]+dp[v][k]+tmp[k]);}}int main(){ //freopen("data.in","r",stdin); //freopen("data.out","w",stdout); int i,j,k,m,n,p; while(~scanf("%d%d",&n,&m)){ memset(head,-1,sizeof(head)); tol=0; for(i=1;i