UVa 208 消防车（dfs+剪枝）

https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=144

题意：给出一个n个结点的无向图以及某个结点k，按照字典序从小到大顺序输出从1到结点k的所有路径。

思路：如果直接矩阵深搜的话是会超时的，所以我们可以从终点出发，将与终点相连的连通块保存起来，这样dfs深搜时可以剪枝掉一些到达不了的点。只要解决了这个，dfs就是小问题。

这道题还有点坑的就是输出格式和它所给的格式不一样，注意一下。

 1 #include<iostream>
 2 #include<cstring>
 3 using namespace std;
 4
 5 const  int maxn = 22;
 6
 7 int n, step, route;
 8 int map[maxn][maxn];
 9 int path[maxn];
10 int vis[maxn];
11 int trunk[maxn];
12
13 void init(int cur)   //从终点开始遍历，保存与终点相连的连通块
14 {
15     trunk[cur] = 1;
16     for (int i = 1; i < maxn; i++)
17     {
18         if (map[cur][i] && !trunk[i])
19             init(i);
20     }
21 }
22
23 void dfs(int cur, int step)
24 {
25     if (cur == n)
26     {
27         cout << "1";
28         for (int i = 1; i < step; i++)
29             cout << " " << path[i];
30         cout << endl;
31         memset(path, 0, sizeof(0));
32         route++;
33     }
34     for (int i = 0; i < maxn; i++)
35     {
36         if (map[cur][i] && !vis[i] && trunk[i])
37         {
38             vis[i] = 1;
39             path[step] = i;
40             dfs(i, step + 1);
41             vis[i] = 0;
42         }
43     }
44     return;
45 }
46
47 int main()
48 {
49     //freopen("D:\\txt.txt", "r", stdin);
50     int a, b, kase = 0;
51     while (cin >> n && n)
52     {
53         memset(vis, 0, sizeof(vis));
54         memset(map, 0, sizeof(map));
55         memset(path, 0, sizeof(path));
56         memset(trunk, 0, sizeof(trunk));
57         while (cin >> a >> b)
58         {
59             if (!a && !b)  break;
60             map[a][b] = map[b][a] = 1;
61         }
62         vis[1] = 1;
63         step = 1;
64         route = 0; //记录路径数量
65         init(n); //计算保存连通块
66         cout << "CASE " << ++kase << ":" << endl;
67         dfs(1, 1);
68         cout << "There are " << route << " routes from the firestation to streetcorner " << n << "." << endl;
69
70     }
71     return 0;
72 }