Codeforces Round #575 (Div. 3)
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。
1 #include<iostream>
2 #include<cstdio>
3 #include<cmath>
4 #include<cstring>
5 #include<map>
6 #include<set>
7 #include<vector>
8 #include<algorithm>
9 #include<queue>
10 #include<unordered_map>
11 #include<list>
12 using namespace std;
13 #define ll long long
14 const int mod=1e9+7;
15 const ll int inf=1e18+7;
16
17 const int maxn=4e5+5;
18
19 typedef struct
20 {
21 int u;
22 int v;
23 ll int w;
24 }St;
25 St sum[maxn];
26
27 bool cmp(const St &a,const St &b)
28 {
29 return a.w < b.w;
30 }
31
32 int n,m,k;
33
34 ll int e[2005][2005];
35
36 void init()
37 {
38 for(int i=0;i<=1505;i++)
39 {
40 for(int j=0;j<=1505;j++)
41 {
42 if(i==j)
43 e[i][j]=0;
44 else
45 e[i][j]=inf;
46 }
47 }
48 }
49
50 int main()
51 {
52 ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
53
54 init();
55
56 cin>>n>>m>>k;
57
58 vector<ll int>v;
59
60 for(int i=0;i<m;i++)
61 cin>>sum[i].u>>sum[i].v>>sum[i].w;
62
63 sort(sum,sum+m,cmp);
64
65 for(int i=0;i<min(m,k);i++)
66 {
67 v.push_back(sum[i].u);
68 v.push_back(sum[i].v);
69 }
70
71 //离散化
72 sort(v.begin(),v.end());//排序
73 v.erase(unique(v.begin(),v.end()),v.end());//去重
74
75 int now=v.size();
76
77 for(int i=0;i<min(m,k);i++)
78 {
79 sum[i].u=lower_bound(v.begin(),v.end(),sum[i].u)-v.begin()+1;//加一防止点从0开始
80 sum[i].v=lower_bound(v.begin(),v.end(),sum[i].v)-v.begin()+1;
81
82 e[sum[i].u][sum[i].v]=sum[i].w;
83 e[sum[i].v][sum[i].u]=sum[i].w;
84
85 }
86
87 for(int k=1;k<=now;k++)
88 {
89 for(int i=1;i<=now;i++)
90 {
91 for(int j=1;j<=now;j++)
92 {
93 if(e[i][k]+e[k][j]<e[i][j])
94 {
95 e[i][j]=e[i][k]+e[k][j];
96 }
97 }
98 }
99 }
100
101 vector<ll int>edge;
102
103 for(int i=1;i<=now;i++)
104 {
105 for(int j=i+1;j<=now;j++)
106 {
107 edge.push_back(e[i][j]);
108 }
109 }
110
111 sort(edge.begin(),edge.end());;
112
113 cout<<edge[k-1]<<endl;
114
115 return 0;
116 }