E. Sonya and Ice Cream（开拓思维）

E. Sonya and Ice Cream

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Sonya likes ice cream very much. She eats it even during programming competitions. That is why the girl decided that she wants to open her own ice cream shops.

Sonya lives in a city with $\mathrm{nn junctions\; and n-1n-1 streets\; between\; them.\; All\; streets\; are\; two-way\; and\; connect\; two\; junctions.\; It\; is\; possible\; to\; travel\; from\; any\; junction\; to\; any\; other\; using\; one\; or\; more\; streets.\; City\; Hall\; allows\; opening\; shops\; only\; on\; junctions.\; The\; girl\; cannot\; open\; shops\; in\; the\; middle\; of\; streets.}$

Sonya has exactly $\mathrm{kk friends\; whom\; she\; can\; trust.\; If\; she\; opens\; a\; shop,\; one\; of\; her\; friends\; has\; to\; work\; there\; and\; not\; to\; allow\; anybody\; to\; eat\; an\; ice\; cream\; not\; paying\; for\; it.\; Since\; Sonya\; does\; not\; want\; to\; skip\; an\; important\; competition,\; she\; will\; not\; work\; in\; shops\; personally.}$

Sonya wants all her ice cream shops to form a simple path of the length $\mathrm{rr (1\le r\le k1\le r\le k),\; i.e.\; to\; be\; located\; in\; different\; junctions f1,f2,\dots ,frf1,f2,\dots ,fr and\; there\; is\; street\; between fifi and fi+1fi+1 for\; each ii from 11 to r-1r-1.}$

The girl takes care of potential buyers, so she also wants to minimize the maximum distance between the junctions to the nearest ice cream shop. The distance between two junctions $\mathrm{aa and bb is\; equal\; to\; the\; sum\; of\; all\; the\; street\; lengths\; that\; you\; need\; to\; pass\; to\; get\; from\; the\; junction aa to\; the\; junction bb.\; So\; Sonya\; wants\; to minimize}$

$${maxamin1\le i\le rda,fimaxamin1\le i\le rda,fi}^{}$$

where $\mathrm{aa takes\; a\; value\; of\; all\; possible nn junctions, fifi —\; the\; junction\; where\; the ii-th\; Sonya\text{'}s\; shop\; is\; located,\; and dx,ydx,y —\; the\; distance\; between\; the\; junctions xx and yy.}$

Sonya is not sure that she can find the optimal shops locations, that is why she is asking you to help her to open not more than $\mathrm{kk shops\; that\; will\; form\; a\; simple\; path\; and\; the\; maximum\; distance\; between\; any\; junction\; and\; the\; nearest\; shop\; would\; be\; minimal.}$

Input

The first line contains two integers $\mathrm{nn and kk (1\le k\le n\le 1051\le k\le n\le 105) —\; the\; number\; of\; junctions\; and\; friends\; respectively.}$

Each of the next $\mathrm{n-1n-1 lines\; contains\; three\; integers uiui, vivi,\; and didi (1\le ui,vi\le n1\le ui,vi\le n, vi\ne uivi\ne ui, 1\le d\le 1041\le d\le 104) —\; junctions\; that\; are\; connected\; by\; a\; street\; and\; the\; length\; of\; this\; street.\; It\; is\; guaranteed\; that\; each\; pair\; of\; junctions\; is\; connected\; by\; at\; most\; one\; street.\; It\; is\; guaranteed\; that\; you\; can\; get\; from\; any\; junctions\; to\; any\; other.}$

Output

Print one number — the minimal possible maximum distance that you need to pass to get from any junction to the nearest ice cream shop. Sonya's shops must form a simple path and the number of shops must be at most $\mathrm{kk.}$

Examples

input

Copy

6 2
1 2 3
2 3 4
4 5 2
4 6 3
2 4 6

output

Copy

4

input

Copy

10 3
1 2 5
5 7 2
3 2 6
10 6 3
3 8 1
6 4 2
4 1 6
6 9 4
5 2 5

output

Copy

7

Note

In the first example, you can choose the path 2-4, so the answer will be 4.

The first example.

In the second example, you can choose the path 4-1-2, so the answer will be 7.

The second example.

 

      求一个树的直径 

　　这个运用了数据结构set优化  太强了 大佬操作  

　　一定要好好学学这个操作　　

#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e5 + 10;
set<pair<int, int> > d[maxn], s;
int n, k, ans = 0;
int main() {
    scanf("%d%d", &n, &k);
    for (int i = 0 ; i < n - 1 ; i++ ) {
        int u, v, w;
        scanf("%d%d%d", &u, &v, &w);
        d[u].insert(make_pair(v, w));
        d[v].insert(make_pair(u, w));
    }
    for (int i = 1 ; i <= n ; i++)
        if (d[i].size() == 1) s.insert(make_pair((*d[i].begin()).second, i));
    while( n > k || s.size() > 2) {
        ans = (*s.begin()).first;
        int i = (*s.begin()).second;
        s.erase(s.begin());
        int next = (*d[i].begin()).first;
        d[next].erase(d[next].lower_bound(make_pair(i, 0)));
        n--;
        if (d[next].size() == 1)
            s.insert(make_pair((*d[next].begin()).second + ans, next));
    }
    printf("%d
", ans);
    return 0;
}