西南交通大学第十三届ACM决赛-重现赛-E（DFS）

题目描述

The term of this problem is the same as the problem "Maximize The Beautiful Value", the only exception — The sequence may not necessary to to non-decreasing.

输入描述:

 The first line contains an positive integer T(1≤T≤10), represents there are T test cases. 
 For each test case: 
 The first line contains two positive integers n,k(1≤n≤105,1≤k<n)，the length of the sequence ,the least steps you need to move. 
 The second line contains n integers a1,a2…an(1≤ai≤108) - the sequence.

输出描述:

For each test case, you should output the max F(n).

示例1



题意：给你n个城市，n-1条边，这n个城市可相互到达，现在要求将这n个城市分成n/2对，
问你这n/2对城市之间的距离之和最小是多少。
题解：考虑贪心，每个点和相邻的点之间的距离一定是最小，呢找完所有相邻点后剩下的点怎么办？
还是一样的，一定是尽量走尽可能少的边，假如说两个点不是相邻的，但是这两个点到他们共同的
父亲的距离之和一定是最优解，因此可以考虑异或模拟这个过程即可。
#include<set>
#include<map>
#include<stack>
#include<queue>
#include<vector>
#include<string>
#include<time.h>
#include<math.h>
#include<stdio.h>
#include<iostream>
#include<string.h>
#include<stdlib.h>
#include<algorithm>
#include<functional>
using namespace std;
#define ll long long
#define inf 1000000000
#define mod 1000000007
#define maxn 100100
#define lowbit(x) (x&-x)
#define eps 1e-9
struct node
{
ll x,y;
};
vector<node>q[maxn];
bool flag[maxn];
ll ans;
void dfs(int u,int p)
{
int i;
for(i=0;i<q[u].size();i++)
{
node v=q[u][i];
if(v.x==p)
continue;
dfs(v.x,u);
if(flag[v.x]==1)
continue;
ans+=v.y;
flag[u]^=1;
flag[v.x]^=1;
}
}
int main(void)
{
ll T,n,i,x,y,z;
scanf("%lld",&T);
while(T--)
{
ans=0;
scanf("%lld",&n);
memset(flag,0,sizeof(flag));
for(i=1;i<n;i++)
{
scanf("%lld%lld%lld",&x,&y,&z);
node tmp;
tmp.x=y;tmp.y=z;
q[x].push_back(tmp);
tmp.x=x;tmp.y=z;
q[y].push_back(tmp);
}
dfs(1,0);
printf("%lld\n",ans);
for(i=1;i<=n;i++)
q[i].clear();
}
return 0;
}