toj 2870 理解dijstra

题意：求离一个城市第k远的城市的编号，无负权边。

思路：根据dijstra算法的特点可以知道，外层循环n次，每次求出的点就是离源点第i远的点的最短路。所以迭代k次，第k次新求出的点即为答案。

 1 #include <algorithm>
 2 #include <iostream>
 3 #include <cstring>
 4 #include <cstdio>
 5 using namespace std;
 6 
 7 const int INF = 9999999;
 8 const int N = 201;
 9 const int M = 20000;
10 int head[N];
11 int dist[N];
12 bool visit[N];
13 int n, m, k, e;
14 
15 struct Edge
16 {
17     int v, next, w;
18 } edge[M];
19 
20 void addEdge( int u, int v, int w )
21 {
22     edge[e].v = v;
23     edge[e].w = w;
24     edge[e].next = head[u];
25     head[u] = e++;
26 }
27 
28 int dij( int s )
29 {
30     memset( visit, false, sizeof(visit) );
31     for ( int i = 0; i <= n; i++ )
32     {
33         dist[i] = INF;
34     }
35     dist[s] = 0;
36     int u;
37     for ( int i = 0; i <= k; i++ )
38     {
39         u = n;
40         for ( int j = 0; j < n; j++ )
41         {
42             if ( !visit[j] && dist[j] < dist[u] )
43             {
44                 u = j;
45             }
46         }
47         visit[u] = true;
48         for ( int j = head[u]; j != -1; j = edge[j].next )
49         {
50             int v = edge[j].v, w = edge[j].w;
51             if ( !visit[v] && dist[u] + w < dist[v] )
52             {
53                 dist[v] = dist[u] + w;
54             }
55         }
56     }
57     return u;
58 }
59 
60 int main ()
61 {
62     while ( scanf("%d", &n), n )
63     {
64         scanf("%d", &m);
65         e = 0;
66         memset( head, -1, sizeof(head) );
67         while ( m-- )
68         {
69             int u, v, w;
70             scanf("%d%d%d", &u, &v, &w);
71             addEdge( u, v, w );
72             addEdge( v, u, w );
73         }
74         scanf("%d", &k);
75         printf("%d\n", dij(0));
76     }
77     return 0;
78 }

 

posted @ 2015-07-22 15:19  hxy_has_been_used  阅读(165)  评论(0)  编辑  收藏  举报