[codeforces1196F]K-th Path

time limit per test : 2.5 seconds
memory limit per test : 256 megabytes

You are given a connected undirected weighted graph consisting of $n$ vertices and $m$ edges.

You need to print the $k$-th smallest shortest path in this graph (paths from the vertex to itself are not counted, paths from $i$ to $j$ and from $j$ to $i$ are counted as one).
More formally, if $d$ is the matrix of shortest paths, where di,j is the length of the shortest path between vertices $i$ and $j (1≤i<j≤n)$, then you need to print the k-th element in the sorted array consisting of all di,j, where $1≤i<j≤n$.
Input

The first line of the input contains three integers $n,m$ and $k (2≤n≤2⋅10^5, n−1≤m≤min(n(n−1)/{2},2⋅10^5), 1≤k≤min(n(n−1)/2,400)$ — the number of vertices in the graph, the number of edges in the graph and the value of $k$, correspondingly.

Then $m$ lines follow, each containing three integers $x$, $y$ and $w (1≤x,y≤n, 1≤w≤10^9, x≠y)$ denoting an edge between vertices $x$ and $y$ of weight $w$.

It is guaranteed that the given graph is connected (there is a path between any pair of vertices), there are no self-loops (edges connecting the vertex with itself) and multiple edges (for each pair of vertices $x$ and $y$, there is at most one edge between this pair of vertices in the graph).

Output

Print one integer — the length of the $k$-th smallest shortest path in the given graph (paths from the vertex to itself are not counted, paths from $i$ to $j$ and from $j$ to $i$ are counted as one).
Examples
Input

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

Output

3

Input

7 15 18
2 6 3
5 7 4
6 5 4
3 6 9
6 7 7
1 6 4
7 1 6
7 2 1
4 3 2
3 2 8
5 3 6
2 5 5
3 7 9
4 1 8
2 1 1

Output

9

题意：
给定一个无向图
要求做出求出两两点之间距离中，第 $k$小的那个
$k<=400$

题解：
考虑将所有现有距离丢到小根堆中
然后每次取出一个点对及其之间距离更新新的距离加入小根堆
一共做 $k$次即可

#include<bits/stdc++.h>
#define ll long long
#define paa pair<ll,pair<int,int> >
using namespace std;
priority_queue<paa,vector<paa>,greater<paa> >q;
map<int,ll>mp[200004];
int n,m,k;
int getu(paa A){
    return A.second.first;
}
int getv(paa A){
    return A.second.second;
}
int main(){
    scanf("%d%d%d",&n,&m,&k);
    while(!q.empty())q.pop();
    for(int i=1;i<=n;i++)mp[i].clear();
    for(int i=1;i<=m;i++){
        int u,v,w;scanf("%d%d%d",&u,&v,&w);
        q.push({1LL*w,{u,v}});
    }
    ll ans=0;
    while(k&&(!q.empty())){
        paa now=q.top();q.pop();
        int u=getu(now),v=getv(now);
        if(mp[u].count(v))continue;
        for(auto x:mp[u]){
            q.push({x.second+now.first,{v,x.first}});
        }
        for(auto x:mp[v]){
            q.push({x.second+now.first,{u,x.first}});
        }
        mp[v][u]=mp[u][v]=now.first;
        ans=now.first;
        k--;
    }
    printf("%lld\n",ans);
    return 0;
}