poj 2449 A*求k短路

A*的入门题目，需要注意的是当图中只有一个点的时候k短路是不存在的。

  1 #include <iostream>
  2 #include <cstring>
  3 #include <cstdio>
  4 #include <queue>
  5 using namespace std;
  6 
  7 const int INF = 0x3f3f3f3f;
  8 const int N = 1001;
  9 const int M = 100000;
 10 int head[N];
 11 int rhead[N];
 12 int dist[N];
 13 int visit[N];
 14 int n, m, e;
 15 
 16 struct Edge
 17 {
 18     int v, next, w;
 19 } edge[M], redge[M];
 20 
 21 void addEdge( int u, int v, int w )
 22 {
 23     edge[e].v = v;
 24     edge[e].w = w;
 25     edge[e].next = head[u];
 26     head[u] = e;
 27     redge[e].v = u;
 28     redge[e].w = w;
 29     redge[e].next = rhead[v];
 30     rhead[v] = e;
 31     e++;
 32 }
 33 
 34 void spfa( int t )
 35 {
 36     memset( dist, 0x3f, sizeof(dist) );
 37     memset( visit, 0, sizeof(visit) );
 38     int q[N], top = 0;
 39     dist[t] = 0;
 40     q[top] = t;
 41     top = ( top + 1 ) % n;
 42     visit[t] = 1;
 43     for ( int i = 0; i != top; i = ( i + 1 ) % n )
 44     {
 45         int u = q[i];
 46         visit[u] = 0;
 47         for ( int j = rhead[u]; j != -1; j = redge[j].next )
 48         {
 49             int v = redge[j].v, w = redge[j].w;
 50             if ( dist[v] > dist[u] + w )
 51             {
 52                 dist[v] = dist[u] + w;
 53                 if ( !visit[v] )
 54                 {
 55                     q[top] = v;
 56                     top = ( top + 1 ) % n;
 57                     visit[v] = 1;
 58                 }
 59             }
 60         }
 61     }
 62 }
 63 
 64 struct Node 
 65 {
 66     int u, g;
 67     Node(){}
 68     Node( int _u, int _g )
 69     {
 70         u = _u, g = _g;
 71     }
 72     bool operator < ( const Node & o ) const 
 73     {
 74         return dist[u] + g > dist[o.u] + o.g;
 75     }
 76 };
 77 
 78 int a_star( int s, int t, int k )
 79 {
 80     priority_queue<Node> q;
 81     q.push( Node( s, 0 ) );
 82     memset( visit, 0, sizeof(visit) );
 83     while ( !q.empty() )
 84     {
 85         Node cur = q.top();
 86         q.pop();
 87         visit[cur.u]++;
 88         if ( visit[t] == k ) return cur.g;
 89         if ( visit[cur.u] > k ) continue;
 90         for ( int i = head[cur.u]; i != -1; i = edge[i].next )
 91         {
 92             int v = edge[i].v, w = edge[i].w;
 93             q.push( Node( v, cur.g + w ) );
 94         }
 95     }
 96     return -1;
 97 }
 98 
 99 int main ()
100 {
101     while ( scanf("%d%d", &n, &m) != EOF )
102     {
103         e = 0;
104         memset( head, -1, sizeof(head) );
105         memset( rhead, -1, sizeof(rhead) );
106         while ( m-- )
107         {
108             int u, v, w;
109             scanf("%d%d%d", &u, &v, &w);
110             addEdge( u, v, w );
111         }
112         int s, t, k;
113         scanf("%d%d%d", &s, &t, &k);
114         spfa(t);
115         if ( dist[s] == INF )
116         {
117             printf("-1\n");
118             continue;
119         }
120         if ( s == t ) k++;
121         printf("%d\n", a_star( s, t, k ));
122     }
123     return 0;
124 }

 

posted @ 2015-08-10 19:26  hxy_has_been_used  阅读(147)  评论(0)  编辑  收藏  举报