每日一练~Shortest Path~解题报告

Shortest Path

题目描述：

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：

题目大意：

有一个地方有n个城市，每个城市之间有一个距离值，求出n/2对的城市之间最短距离之和。

思路分析：

这道题一开始的时候错误认为是直接从把每一个节点当作父子点，然后去搜索每一个节点，从子节点最短距离来判断，当找到最短就标记，然后就直接TL了，数据太大了。应该这样去看，用前向星链式去做各个点的边，题目给的是双向带权值图，所以做边的时候，要做双向。寻找最短距离一般都是在父节点和兄弟节点去寻找，然后当一个节点存在不是偶数边的时候，说明这个点需要与其他节点配对。

代码：

#include
#include
#include
#include
#include
#include
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);
	}
}