Dijkstra算法

 

 1 #include <iostream>
 2 #include <cstdlib>
 3 #include <cstring>
 4 #include <map>
 5 using namespace std;
 6 /*==================================================*\
 7  | Dijkstra 数组实现O （N^2 ）
 8  | Dijkstra --- 数组实现( 在此基础上可直接改为STL 的Queue实现)
 9  | lowcost[] --- beg 到其他点的最近距离
10  | path[] -- beg为根展开的树，记录父亲结点
11  \*==================================================*/
12 #define INF 0x03F3F3F3F
13 //const int N;
14 map<int, int> path, vis;
15 void Dijkstra(map<int, map<int, int> > cost, map<int, int>& lowcost, int n, int beg) {
16     int i, j, min;
17 //    memset(vis, 0, sizeof(vis));
18     vis[beg] = 1;
19     for (i = 0; i < n; i++) {
20         lowcost[i] = cost[beg][i];
21         path[i] = beg; //最大的距离
22     }
23     lowcost[beg] = 0; //起点到自身的距离肯定是0
24     path[beg] = -1; //  树根的标记
25     int pre = beg;
26     for (i = 1; i < n; i++) {
27         min = INF;
28         for (j = 0; j < n; j++)
29             // 下面的加法可能导致溢出,INF不能取太大
30             //不能有负值，否则会出问题
31             if (vis[j] == 0 && lowcost[pre] + cost[pre][j] < lowcost[j]) {
32                 lowcost[j] = lowcost[pre] + cost[pre][j];
33                 path[j] = pre;
34             }
35         for (j = 0; j < n; j++)
36             if (vis[j] == 0 && lowcost[j] < min) { //选择下一个扩展的点
37                 min = lowcost[j];
38                 pre = j;
39             }
40         vis[pre] = 1;
41     }
42 }
43 int main() {
44     map<int, map<int, int> > cost;
45     map<int, int> lowcost;
46     for (int i = 0; i < 4; i++)
47         for (int j = 0; j < 4; j++) {
48             if (i != j)
49                 cost[i][j] = INF;
50             else
51                 cost[i][j] = 0;
52         }
53     cost[0][1] = 3;
54     cost[0][2] = 1;
55     cost[0][3] = 10;
56     cost[2][1] = 1;
57     cost[2][3] = 1;
58 
59     for (int i = 0; i < 4; i++)
60         vis[i]=0;
61     Dijkstra(cost,lowcost, 4,0);
62     for (int i = 0; i < 4; i++)
63             cout<<lowcost[i]<<endl;
64     return 0;
65 }

 

posted on 2012-07-25 00:15  kakamilan  阅读(156)  评论(0)    收藏  举报

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