「CF1196F」K-th Path【floyd】

最新推荐文章于 2022-06-10 22:40:58 发布

wzw1105

最新推荐文章于 2022-06-10 22:40:58 发布

阅读量727

点赞数

CC 4.0 BY-SA版权

本文链接：https://blog.youkuaiyun.com/qq_40791842/article/details/97315009

离散化同时被 2 个专栏收录

4 篇文章

订阅专栏

最短路径

4 篇文章

订阅专栏

本文介绍了一种求解图中所有点对的最短路径第K小的算法，通过预处理前K小的边进行路径压缩，再对所有可能的最短路径进行排序，最后输出第K小的路径长度。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

F. K-th Path

time limit per test 2.5 seconds memory limit per test 256 megabytes input standard input output standard output

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

You need to print the $k - t h$ 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 $d_{i,j}$ is the length of the shortest path between vertices $i$ and $j$ $(1≤i<j≤n)(1\leq i<j \leq n)$ , then you need to print the $k - t h$ element in the sorted array consisting of all $d_{i,j}$ , where $1≤i<j≤n1\leq i<j\leq n$ .

Input

The first line of the input contains three integers $n, m$ and $k$ $(2≤n≤2⋅105,n−1≤m≤min(n×(n−1)2,2×105),1≤k≤min(n×(n−1)2,400)(2\leq n\leq 2⋅10^5, n−1\leq m\leq min(\frac{n\times (n−1)}{2},2 \times10^5), 1\leq k\leq min(\frac{n\times (n−1)}{2},400)$ — 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 $w$ $(1≤x,y≤n,1≤w≤109,x!=y)(1\leq x,y \leq n, 1\leq w \leq 10^9, x !=y)$ 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 $k - t h$ 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

output

input

output

题意

给你一张图，求所有点对的最短路径第 $k$ 小， $u$ 到 $v$ 最短路径路径与 $v$ 到 $u$ 路径只算一次， $k≤400k\leq 400$

题解

看到 $k$ 这么小，自然想到暴力搞，对！只需要取出前 $k$ 小的边进行路径压缩就行了，然后暴力将这些所有最短路径排序，输出第 $k$ 大，至于为什么不需要第 $k + 1$ 大的边呢？都第 $k + 1$ 大了还要它干嘛？

代码

 #include<bits/stdc++.h>
 
using namespace std;
const int maxn=5e5+10;
#define inf 0x3f3f3f3f3f3f3f3f
 
long long dp[805][805],dis[maxn];
int tot=0,n,m,k,b[maxn],cnt=0;
 
struct node{
    int u,v,w;
    friend bool operator<(const node&a,const node&b){
        return a.w<b.w;
    }
}a[maxn];
 
int main()
{
    scanf("%d %d %d",&n,&m,&k);
    for(int i=1;i<=m;i++) scanf("%d %d %d",&a[i].u,&a[i].v,&a[i].w);
    sort(a+1,a+m+1);memset(dp,0x3f,sizeof(dp));
    for(int i=1;i<=min(k,m);i++) b[++cnt]=a[i].u,b[++cnt]=a[i].v;
    sort(b+1,b+cnt+1);int qwq=unique(b+1,b+cnt+1)-b-1;
    for(int i=1;i<=min(k,m);i++) {
        int from=lower_bound(b+1,b+qwq+1,a[i].u)-b,to=lower_bound(b+1,b+qwq+1,a[i].v)-b;
        dp[from][to]=dp[to][from]=a[i].w;
    }
    for(int k=1;k<=qwq;k++) for(int i=1;i<=qwq;i++) for(int j=1;j<=qwq;j++)if(dp[i][k]+dp[k][j]<dp[i][j]) dp[i][j]=dp[i][k]+dp[k][j];
    for(int i=1;i<=qwq;i++) for(int j=i+1;j<=qwq;j++) if(dp[i][j]!=inf) dis[++tot]=dp[i][j];
    sort(dis+1,dis+tot+1);
    printf("%lld\n",dis[k]);    
}