题意

给出一个无向图和终点的编号
按字典序枚举出从1到终点的路径

思路

要事先判断结点1是否可以到达结点k, 用一个bool judge()函数判断一下从终点能否回到1点即可. 如果无解直接输出有0种走法
有解则用dfs即可. 因为每种走法里一个编号只能走一次, 用一个数组m[]标记是否走过. 记得每次标记之后进入dfs(), 出来之后要清除该标记, 以防影响程序后面的判断
DFS在回溯时要取消原先的标记
而BFS不存在回溯也就不存在取消标记这一问题。

话说这道题的格式, 是没有前导空格的
VJ给的PDF显示有点问题 =_=

AC代码

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>using namespace std;#define mst(a) memset(a,0,sizeof(a));
const int maxn = 30;
int G[maxn][maxn], vis[maxn][maxn], m[maxn], r[maxn];
int aim, mmax, all;bool judge(int n)
{if( n == 1 )    return true;m[n] = 1;for( int i = mmax; i >= 1; i-- ){if( m[i] )    continue;if( G[n][i] )if( judge(i) )return true;}return false;
}void print( int len )
{all++;printf("1");for( int i = 0; i < len; i++ )printf(" %d",r[i]);puts("");
}void dfs( int n, int k )
{if( n == aim )print(k);for( int i = 2; i <= mmax; i++ ){if( !m[i] && G[n][i] ){m[i] = 1;r[k] = i;dfs(i, k+1);m[i] = 0;}}
}void solve()
{mst(m);dfs(1, 0);
}int main()
{int a, b, kase = 0;while( ~scanf("%d",&aim) ){mmax = aim, all = 0;mst(G);mst(vis);mst(m);mst(r);while( scanf("%d%d",&a,&b) == 2 && a ){G[a][b] = 1;G[b][a] = 1;    //双向可走mmax = max(mmax,a);mmax = max(mmax,b);}printf("CASE %d:\n",++kase);if(!judge(aim)){printf("There are 0 routes from the firestation to streetcorner %d.\n",aim);continue;}solve();printf("There are %d routes from the firestation to streetcorner %d.\n",all, aim);}return 0;
}