CCF 201703-4 地铁修建（最小瓶颈路）

题意：读清楚之后发现就是求最小瓶颈路的瓶颈，而容易知道求最小瓶颈路就是求最小生成树的最大边

 

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<algorithm>
 4 #include<iostream>
 5 #define LL long long
 6 #define debug(x) cout << "[" << x << "]" << endl
 7 using namespace std;
 8 
 9 const int mx = 1e5+10;
10 
11 struct edge{
12     int u, v, w;
13     edge(int u = 0, int v = 0, int w = 0): u(u), v(v), w(w){}
14     bool operator < (const edge& a) const {
15         return w < a.w;
16     }
17 }e[mx*2];
18 int pa[mx];
19 
20 int fx(int x){
21     return pa[x] == x ? x : pa[x] = fx(pa[x]);
22 }
23 
24 int main(){
25     int n, m, u, v, w;
26     scanf("%d%d", &n, &m);
27     for (int i = 0; i < m; i++){
28         scanf("%d%d%d", &u, &v, &w);
29         e[i] = edge(u, v, w);
30     }
31     for (int i = 1; i <= n; i++) pa[i] = i;
32     sort(e, e+m);
33     int ans = 0;
34     for (int i = 0; i < m; i++){
35         int a = fx(e[i].u), b = fx(e[i].v);
36         if (a != b){
37             pa[a] = b;
38             ans = max(ans, e[i].w);
39         }
40         if (fx(1) == fx(n)) break;
41     }
42     printf("%d\n", ans);
43     return 0;
44 }

 

posted @ 2018-09-15 17:45  QAQorz  阅读(221)  评论(0)  编辑  收藏  举报