题意:按最小字典序输出a到b 的所有路径。
思路:先处理出个点到目标点b的情况(是否能到达),搜索即可。
最开始我只判了a能否到b,然后给我的是WA,然后看了半天感觉思路没什么问题,然后把所有点都处理出来,AC
实在是看不懂- -,好无语。
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <queue>
#include <algorithm>
typedef long long ll;
using namespace std;
int tmap[200][200];
const int inf = 0x3f3f3f3f;
int ans[200];
int can[200];
int n,a,b;
int tmax;
int vis[200];
int all; void dfs(int cur,int num)
{
if(cur == n)
{
printf("1");
for(int i = 1; i < num; i++)
printf(" %d",ans[i]);
printf("\n");
all++;
return ;
}
for(int i = 1; i <= tmax; i++)
{
if(!vis[i] && tmap[cur][i] && can[i])
{
vis[i] = 1;
ans[num] = i;
dfs(i,num+1);
vis[i] = 0;
}
}
return ;
} void get_can(int cur) //是否能到达终点
{
can[cur] = 1;
for(int i =1 ; i <= tmax; i++)
if(tmap[cur][i] && !can[i])
get_can(i);
} int main()
{
int cas = 1;
while(scanf("%d",&n)!=EOF)
{
memset(vis,0,sizeof(vis));
memset(can,0,sizeof(can));
memset(tmap,0,sizeof(tmap));
while(scanf("%d %d",&a,&b))
{
if(a == 0 && b == 0)
break;
tmap[a][b] = tmap[b][a] = 1;
if(a > tmax)
tmax = a;
if(b > tmax)
tmax = b;
}
get_can(n);
printf("CASE %d:\n",cas++);
if(!can[n])
{
printf("There are 0 routes from the firestation to streetcorner %d.\n",n);
continue;
} vis[1] = 1;
all = 0;
dfs(1,1);
printf("There are %d routes from the firestation to streetcorner %d.\n",all,n);
}
return 0;
}