HDU-5102-The K-th Distance【思维】【好题】

Problem Description

Given a tree, which has n node in total. Define the distance between two node u and v is the number of edge on their unique route. So we can have n(n-1)/2 numbers for all the distance, then sort the numbers in ascending order. The task is to output the sum of the first K numbers.

Input

There are several cases, first is the number of cases T. (There are most twenty cases).

For each case, the first line contain two integer n and K (2≤n≤100000,0≤K≤min(n(n−1)/2,106) ). In following there are n-1 lines. Each line has two integer u , v. indicate that there is an edge between node u and v.

Output

For each case output the answer.

Sample Input

2

3 3

1 2

2 3

5 7

1 2

1 3

2 4

2 5

Sample Output

4

10

题目链接：HDU-5102

题目大意：给出一棵树，每条边的权值为1，将树上所有两点之间的距离存进数组，进行从小到大排序，问前k个的值为多少。

题目思路：因为k的范围为10^6,所以可以从这里入手。

用一个队列，q:

u，v: 左边界限，右边界限
w: 层数（左界限到右界限的距离）

因为得到的数组是从小到大排序，所以我们可以考虑，处理所有长度为1的，然后将左界限和右界限进行扩展一格，处理长度为2的。以此。

注意：刚开始，我为了处理该左界限和右界限是否被计算过，使用了map来存，MLE了。

实际上，计算时，每个值重复两次，所以答案／2就可以了，不需要使用map来标记

参考博客：here

以下是代码：

#include <iostream>
#include <iomanip>
#include <fstream>
#include <sstream>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <cctype>
#include <algorithm>
#include <functional>
#include <numeric>
#include <string>
#include <set>
#include <map>
#include <stack>
#include <vector>
#include <queue>
#include <deque>
#include <list>
typedef long long ll;
using namespace std;
#define maxn 100001
vector <int> vec[maxn];
struct node
{
int u,v,w;
node(int u,int v,int w):u(u),v(v),w(w){}
};
queue<node> q;
map<long long, bool> mp;
int main()
{
int _;
cin >> _;
while(_--)
{
while(!q.empty()) q.pop();
for (int i = 0; i < maxn; i++) vec[i].clear();

int n,k;
cin >> n >> k;
k *= 2;

for (int i = 0; i < n - 1; i++)
{
int u,v;
cin >> u >> v;
vec[u].push_back(v);
vec[v].push_back(u);
}
for (int i = 1; i <= n; i++)
{
q.push(node(i,i,0));
}
int cnt = 0;
long long ans = 0;

while(!q.empty())
{
node front = q.front();
q.pop();

if (cnt >= k) break;

int u = front.u;
int w = front.w;

for (int i = 0; i < vec[u].size(); i++)
{
int v = vec[u][i];
if (v == front.v) continue;

if (cnt < k)
{
cnt++;
ans += w + 1;
q.push(node(v,u,w + 1));
}
}

}
cout << ans / 2 << endl;

}
return 0;
}