旅行商问题(tsp)之分支定界法

http://soj.sysu.edu.cn/show_problem.php?pid=1001&cid=1816

做了一个晚上的题，真是弱爆了...

其实就是深搜最短路，不过加了一个upper bound用来剪枝，因为数据比较小可以过！

深搜还是要熟悉啊！

 1 #include <bits/stdc++.h>
 2 
 3 using namespace std;
 4 
 5 double graph[25][25];
 6 int vis[25];
 7 double a[25];
 8 double ub;
 9 double ans;
10 
11 
12 struct point {
13     int x;
14     int y;
15 }p[25];
16 
17 double dis(point a, point b)
18 {
19     return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
20 }
21 
22 void dfs(int n, int x, double temp, int cnt)
23 {
24 //    printf("%.2lf
", temp);
25     if(cnt == n-2)
26     {
27         ub = min(ub, temp+graph[x][n-1]);
28         return ;
29     }
30     vis[x]=1;
31     for(int i=0; i<n-1; i++)
32     {
33         if(graph[x][i] && !vis[i])
34         {
35             temp += graph[x][i];
36             if(temp < ub)
37                 dfs(n, i, temp, cnt+1);    
38             temp -= graph[x][i];
39             vis[i]=0;    
40                     
41         }
42     }
43     
44 
45 }
46 
47 
48 int main()
49 {
50     int t, n, x, y;
51     scanf("%d", &t);
52     while(t--)
53     {
54         int cnt=0;
55         ub=0x3f3f3f3f;
56         double temp=0;
57         memset(graph, 0, sizeof(graph));
58         memset(vis, 0, sizeof(vis));
59         
60         scanf("%d", &n);
61         for(int i=0; i<n; i++)
62             scanf("%d%d", &p[i].x, &p[i].y);
63 
64         for(int i=0; i<n; i++)
65             for(int j=0; j<n; j++)
66                 graph[i][j] = dis(p[i], p[j]);    
67                         
68         vis[0]=1;
69         dfs(n,0,temp,0);
70         printf("%.2lf
", ub);
71     }
72     return 0;
73 }