lightoj 1002 最小生成树

题意,给出若干条道路(m条)和若干个城市(n个,编号从0~n-1),给出一个起步城市k( 0<=k<=n) , 问所有其他城市到起步城市k的路径中,路径中最长的道路,最短可以 是多少。

分析,虽然一开始看错了题目,当作了最短路做,改了回来,用了并查集+队列拓展。但是Gealo说可以用dij做,数组保存的不是到某点的最短距离,而是到某点的最长的道路长为多少。好像也是可以的。

总之写的是最小生成树,那就用最小生成树写,最小生成树可以完美保证题意,题意中要找到路径,也就是要联通,且最小,符合最小生成树定义。

 1 /* When all else is lost the future still remains. */
 2 #define rep(X,Y,Z) for(int X=(Y);X<(Z);X++)
 3 #define drep(X,Y,Z) for(int X=(Y);X>=(Z);X--)
 4 #define fi first
 5 #define se second
 6 //head
 7 #include <iostream>
 8 #include <stdio.h>
 9 #include <queue>
10 #include <algorithm>
11 using namespace std;
12 #define MAXM 16010
13 #define MAXN 510
14 #define INF 1000000000
15 //int road_val[MAXN][MAXN];
16 pair<int,pair<int,int> > road_val[MAXM];
17 int dis[MAXN][MAXN];
18 int fat[MAXN];
19 int maxdis[MAXN];
20 int dij_dis[MAXN];
21 bool mark[MAXN];
22 void init(){
23     rep(i,0,MAXN) rep(j,0,MAXN) dis[i][j] = INF;
24     rep(i,0,MAXN) fat[i] = i;
25     rep(i,0,MAXN) dij_dis[i] = INF;
26     rep(i,0,MAXN) maxdis[i] = -1;
27     rep(i,0,MAXN) mark[i] = 0;
28     return ;
29 }
30 int find_fa(int x){
31     return fat[x] = (fat[x] == x) ? x : find_fa(fat[x]);
32 }
33 void dij(int pos , int n){
34     queue<int> Q;
35     Q.push(pos);
36     maxdis[pos] = 0;
37     while(!Q.empty()){
38         int now = Q.front();
39         Q.pop();
40         mark[now] = 1;
41         rep(i,0,n){
42             if(mark[i]) continue;
43             if(dis[now][i] >= INF) continue;
44             maxdis[i] = max(maxdis[i],maxdis[now]);
45             maxdis[i] = max(maxdis[i],dis[now][i]);
46             Q.push(i);
47         }
48     }
49     return ;
50 }
51 int main(){
52     int T;
53     scanf("%d",&T);
54     rep(ca,1,T+1){
55         int n , m;
56         scanf("%d %d",&n,&m);
57         init();
58         rep(i,0,m){
59             int u , v , w;
60             scanf("%d %d %d",&u,&v,&w);
61             road_val[i] = make_pair(w,make_pair(u,v));
62         }
63         sort(road_val,road_val+m);
64         rep(i,0,m){
65             int val = road_val[i].fi;
66             int a = road_val[i].se.fi;
67             int b = road_val[i].se.se;
68             int fa = find_fa(a);
69             int fb = find_fa(b);
70             if(fa == fb) continue;
71             fat[fa] = min(fa,fb);
72             fat[fb] = min(fa,fb);
73             dis[a][b] = min(dis[a][b],val);
74             dis[b][a] = min(dis[b][a],val);
75         }
76         int start;
77         scanf("%d",&start);
78         dij(start,n);
79         printf("Case %d:
",ca);
80         rep(i,0,n){
81             if(maxdis[i] < 0) printf("Impossible
");
82             else printf("%d
", maxdis[i]);
83         }
84 
85     }
86     return 0;
87 }
原文地址:https://www.cnblogs.com/ticsmtc/p/5940730.html