Shortest Path
題目描述：
Today HH becomes a designer, and he faces a problem so he asks you for help.
Treeisland is a country with n cities and n−1 two-way road and from any city you can go to any other cities.
HH the designer is going to design a plan to divide n city into n/2 pairs so that the sum of the length between the n/2 pairs city is minimum.
Now HH has finished it but he doesn’t know whether it’s true so he ask you to calculate it together.
It’s guaranteed that n is even.
Input：
The first line contains an positive integer T(1≤T≤100), represents there are T test cases.
For each test case: The first line contains an positive integer n(1≤n≤104), represents the number of cities in Treeisland, it’s guarantee that n is even.
Then n−1 lines followed. Each line contains three positive integer u, v and len, (u≠v,1≤u≤n,1≤v≤n,1≤len≤109)indicating there is a road of length len between u and v.
It’s guarantee you can get to any city from any city.
Output：
For each test case, output in one line an integer, represent the minimum sum of length.
Sample Input：
2
4
1 2 5
2 3 8
3 4 6
6
1 3 5
3 2 3
4 5 4
4 3 9
4 6 10
Sample Output：
11
31
Hint：
題目大意：
有一個地方有n個城市，每個城市之間有一個距離值，求出n/2對的城市之間最短距離之和。
思路分析：
這道題一開始的時候錯誤認爲是直接從把每一個節點當作父子點，然後去搜索每一個節點，從子節點最短距離來判斷，當找到最短就標記，然後就直接TL了，數據太大了。應該這樣去看，用前向星鏈式去做各個點的邊，題目給的是雙向帶權值圖，所以做邊的時候，要做雙向。尋找最短距離一般都是在父節點和兄弟節點去尋找，然後當一個節點存在不是偶數邊的時候，說明這個點需要與其他節點配對。
代碼：
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<vector>
#include<queue>
#include<cstring>
using namespace std;
typedef long long ll;
const int maxn=3e5+6;
typedef struct node
{
int to;
int u;
ll w;
}Lemon;
Lemon Frist[maxn];
ll head[maxn];
long long size[maxn];
ll cnt;
ll ans;
ll read()
{
long long x=0;
long long f=1;
char c1=getchar();
while(c1<'0' || c1>'9')
{
if(c1=='-')
{
f=-1;
}
c1=getchar();
}
while(c1>='0' && c1<='9')
{
x=x*10+c1-'0';
c1=getchar();
}
return x*f;
}
void LemonADD(int u,int v,int z)
{
Frist[cnt].u=head[u];
Frist[cnt].to=v;
Frist[cnt].w=z;
head[u]=cnt++;
}
void LemonDFS(int u,int v,int z)
{
size[u]=1;
for(int i=head[u];i;i=Frist[i].u)
{
int h=Frist[i].to;
if(h==v)continue;
LemonDFS(Frist[i].to,u,Frist[i].w);
size[u]+=size[h];
}
if(size[u]&1)ans+=z;
}
int main()
{
int t;
cin >> t;
while(t--)
{
cnt=1;
ans=0;
memset(head,0,sizeof(head));
long long n=read();
ll z;
int x,y;
for(ll i=0;i<n-1;i++)
{
cin >> x >> y >> z;
LemonADD(x,y,z);
LemonADD(y,x,z);
}
LemonDFS(1,0,0);
printf("%lld\n",ans);
}
}