poj 2253 dijstra变形

dijstra算法的变形，定义：dist[i]为源点到点i的若干路径上的边的最大值的最小值，然后会发现可以用和dijstra一样的贪心方法，当前dist最小的以后都不会再被更新。

 1 #include <iostream>
 2 #include <cstring>
 3 #include <cstdio>
 4 #include <cmath>
 5 using namespace std;
 6
 7 const int INF = 9999999;
 8 const int N = 201;
 9 int head[N];
10 int dist[N];
11 bool visit[N];
12 int n, e;
13
14 struct Node
15 {
16     int x, y;
17 } node[N];
18
19 struct Edge
20 {
21     int v, next, w;
22 } edge[N * N];
23
24 void addEdge( int u, int v, int w )
25 {
26     edge[e].v = v;
27     edge[e].w = w;
28     edge[e].next = head[u];
29     head[u] = e++;
30 }
31
32 int dis( int i, int j )
33 {
34     return ( node[i].x - node[j].x ) * ( node[i].x - node[j].x ) +
35            ( node[i].y - node[j].y ) * ( node[i].y - node[j].y );
36 }
37
38 void dij( int s )
39 {
40     memset( visit, false, sizeof(visit) );
41     for ( int i = 0; i <= n; i++ )
42     {
43         dist[i] = INF;
44     }
45     dist[s] = 0;
46     for ( int i = 1; i <= n; i++ )
47     {
48         int u = 0;
49         for ( int j = 1; j <= n; j++ )
50         {
51             if ( !visit[j] && dist[j] < dist[u] )
52             {
53                 u = j;
54             }
55         }
56         visit[u] = true;
57         for ( int j = head[u]; j != -1; j = edge[j].next )
58         {
59             int v = edge[j].v, w = edge[j].w;
60             if ( visit[v] ) continue;
61             dist[v] = min( dist[v], max( dist[u], w ) );
62         }
63     }
64 }
65
66 int main ()
67 {
68     int _case = 1;
69     while ( scanf("%d", &n), n )
70     {
71         e = 0;
72         memset( head, -1, sizeof(head) );
73         for ( int i = 1; i <= n; i++ )
74         {
75             scanf("%d%d", &node[i].x, &node[i].y);
76             for ( int j = i - 1; j > 0; j-- )
77             {
78                 int tmp = dis( i, j );
79                 addEdge( i, j, tmp );
80                 addEdge( j, i, tmp );
81             }
82         }
83         dij(1);
84         double ans = sqrt( dist[2] * 1.0 );
85         printf("Scenario #%d\nFrog Distance = %.3lf\n\n", _case++, ans);
86     }
87     return 0;
88 }