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。

 #include<iostream>
 #include<cstring>
 using namespace std;

 const  int maxn = ;

 int n, step, route;
 int map[maxn][maxn];
 int path[maxn];
 int vis[maxn];
 int trunk[maxn];

 void init(int cur)   //从终点开始遍历，保存与终点相连的连通块
 {
     trunk[cur] = ;
     for (int i = ; i < maxn; i++)
     {
         if (map[cur][i] && !trunk[i])
             init(i);
     }
 }

 void dfs(int cur, int step)
 {
     if (cur == n)
     {
         cout << "";
         for (int i = ; i < step; i++)
             cout << " " << path[i];
         cout << endl;
         memset(path, , sizeof());
         route++;
     }
     for (int i = ; i < maxn; i++)
     {
         if (map[cur][i] && !vis[i] && trunk[i])
         {
             vis[i] = ;
             path[step] = i;
             dfs(i, step + );
             vis[i] = ;
         }
     }
     return;
 }

 int main()
 {
     //freopen("D:\\txt.txt", "r", stdin);
     int a, b, kase = ;
     while (cin >> n && n)
     {
         memset(vis, , sizeof(vis));
         memset(map, , sizeof(map));
         memset(path, , sizeof(path));
         memset(trunk, , sizeof(trunk));
         while (cin >> a >> b)
         {
             if (!a && !b)  break;
             map[a][b] = map[b][a] = ;
         }
         vis[] = ;
         step = ;
         route = ; //记录路径数量
         init(n); //计算保存连通块
         cout << "CASE " << ++kase << ":" << endl;
         dfs(, );
         cout << "There are " << route << " routes from the firestation to streetcorner " << n << "." << endl;

     }
     return ;
 }