Codeforces Round #575 (Div. 3) F - K-th Path

Codeforces Round #575 (Div. 3)

 

F - K-th Path

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 jto 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≤109, 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：题目意思是给你好多边组成一个图，然后问你任意两点之间第K短路是多少。

思路：这里有个优化，可以先对所有边根据边权排个序，如果总边数超过k，只取前k条边建图就好了，

因为求第k短路，后面的边肯定用不上，（可以自己先思考一下）。然后再离散化一下，拿边跑一遍floyd（k最大400，o（n^3）能接受）。

再暴力把所有边取出来，排个序输出第k短路就好了，这是真的暴力，能过还行，所以说要敢于尝试。

 

 

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<map>
#include<set>
#include<vector>
#include<algorithm>
#include<queue>
#include<unordered_map>
#include<list>
using namespace std;
#define ll long long 
const int mod=1e9+7;
const ll int inf=1e18+7;
 
const int maxn=4e5+5;
 
typedef struct
{
    int u;
    int v;
    ll int w;
}St;
St sum[maxn];
 
bool cmp(const St &a,const St &b)
{
    return a.w < b.w;
}
 
int n,m,k;
 
ll int e[2005][2005];
 
void init()
{
    for(int i=0;i<=1505;i++)
    {
        for(int j=0;j<=1505;j++)
        {
            if(i==j)
                e[i][j]=0;
            else
                e[i][j]=inf;
        }
    }
}
 
int main()
{
    ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
    
    init();
    
    cin>>n>>m>>k;
    
    vector<ll int>v;
    
    for(int i=0;i<m;i++)
        cin>>sum[i].u>>sum[i].v>>sum[i].w;
    
    sort(sum,sum+m,cmp);
    
    for(int i=0;i<min(m,k);i++)
    {
        v.push_back(sum[i].u);
        v.push_back(sum[i].v);
    }
    
    //离散化 
    sort(v.begin(),v.end());//排序 
    v.erase(unique(v.begin(),v.end()),v.end());//去重 
    
    int now=v.size();
    
    for(int i=0;i<min(m,k);i++)
    {
        sum[i].u=lower_bound(v.begin(),v.end(),sum[i].u)-v.begin()+1;//加一防止点从0开始 
        sum[i].v=lower_bound(v.begin(),v.end(),sum[i].v)-v.begin()+1;
        
        e[sum[i].u][sum[i].v]=sum[i].w;
        e[sum[i].v][sum[i].u]=sum[i].w; 
        
    }
    
    for(int k=1;k<=now;k++)
    {
        for(int i=1;i<=now;i++)
        {
            for(int j=1;j<=now;j++)
            {
                if(e[i][k]+e[k][j]<e[i][j])
                {
                    e[i][j]=e[i][k]+e[k][j];
                }
            }
        }
    }
    
    vector<ll int>edge;
    
    for(int i=1;i<=now;i++)
    {
        for(int j=i+1;j<=now;j++)
        {
            edge.push_back(e[i][j]);
        }
    }
    
    sort(edge.begin(),edge.end());;
    
    cout<<edge[k-1]<<endl;
    
    return 0;
}