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 = 22;

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] = 1;
    for (int i = 1; i < maxn; i++)
    {
        if (map[cur][i] && !trunk[i])
            init(i);
    }
}

void dfs(int cur, int step)
{
    if (cur == n)
    {
        cout << "1";
        for (int i = 1; i < step; i++)
            cout << " " << path[i];
        cout << endl;
        memset(path, 0, sizeof(0));
        route++;
    }
    for (int i = 0; i < maxn; i++)
    {
        if (map[cur][i] && !vis[i] && trunk[i])
        {
            vis[i] = 1;
            path[step] = i;
            dfs(i, step + 1);
            vis[i] = 0; 
        }
    }
    return;
}

int main()
{
    //freopen("D:\txt.txt", "r", stdin);
    int a, b, kase = 0;
    while (cin >> n && n)
    {
        memset(vis, 0, sizeof(vis));
        memset(map, 0, sizeof(map));
        memset(path, 0, sizeof(path));
        memset(trunk, 0, sizeof(trunk));
        while (cin >> a >> b)
        {
            if (!a && !b)  break;
            map[a][b] = map[b][a] = 1;
        }
        vis[1] = 1;
        step = 1;
        route = 0; //记录路径数量
        init(n); //计算保存连通块
        cout << "CASE " << ++kase << ":" << endl;
        dfs(1, 1);
        cout << "There are " << route << " routes from the firestation to streetcorner " << n << "." << endl;

    }
    return 0;
}