DFS深度优先搜索

核心伪代码就2行

DFS(this)gray thisDFS all white adj

用DFS打印图

输入

输出

1 2 3 4 5 7 8 6

代码

//DFS.c
# include<stdio.h>
# include<stdlib.h>
# include<stdbool.h>
# include<string.h>
FILE *IN,*OUT;# define N 9int v_num,e_num,first,last,G[N][N],source,color[N];void DFS(int cur){color[cur] = 1; fprintf(OUT,"%d ",cur);for(int i = first; i <= last; i++)if(!color[i] && G[cur][i])DFS(i);
}int main(){IN = fopen("my_input.txt", "r"); OUT = fopen("my_output.txt", "w");//fscanf(IN,"%d %d %d", &v_num,&e_num,&source);first = 1; last = first + v_num - 1;memset(G,0,sizeof(G));while(e_num--){int v1,v2; fscanf(IN,"%d %d", &v1,&v2);G[v1][v2] = G[v2][v1] = 1;}memset(color,0,sizeof(color));DFS(source); fprintf(OUT,"\n");//fclose(IN); fclose(OUT);return 0;
}

可以看到，哪怕写成C语言，DFS函数的body部分还是只有两行，优雅而强大。

用P数组记录前驱pre/path

我特意把8号节点设置为了孤立节点，然后进行DFS，测试能否从8号节点走到source

//DFS_PATH.c
# include<stdio.h>
# include<stdlib.h>
# include<stdbool.h>
# include<string.h>
FILE *IN,*OUT;# define N 9int v_num,e_num,first,last,G[N][N],source;
int P[N];void DFS(int cur){fprintf(OUT,"%d ",cur);for(int i = first; i <= last; i++)if(P[i]==-1 && G[cur][i]){P[i] = cur;DFS(i);}
}bool ifConnectToSource(int cur){if(cur == source)return true;if(cur == -1)return false;return ifConnectToSource(P[cur]);
}int main(){IN = fopen("my_input.txt", "r"); OUT = fopen("my_output.txt", "w");//fscanf(IN,"%d %d %d", &v_num,&e_num,&source);first = 1; last = first + v_num - 1;memset(G,0,sizeof(G));while(e_num--){int v1,v2; fscanf(IN,"%d %d", &v1,&v2);G[v1][v2] = G[v2][v1] = 1;}memset(P,-1,sizeof(P));DFS(source); fprintf(OUT,"\n");for(int i = first; i <= last; i++)fprintf(OUT,"%d\t%d\n",i,P[i]);fprintf(OUT,"%d\n",ifConnectToSource(8));//fclose(IN); fclose(OUT);return 0;
}

每次一次DFS都会消耗一定的能量，假如能量消耗完了就停止DFS

//DFS_ENERGY.c
# include<stdio.h>
# include<stdlib.h>
# include<stdbool.h>
# include<string.h>
FILE *IN,*OUT;# define N 9int v_num,e_num,first,last,G[N][N],source;
int P[N];
int energy;bool DFS(int cur){if(energy == 0)return false;energy--;fprintf(OUT,"%d ",cur);for(int i = first; i <= last; i++)if(P[i]==-1 && G[cur][i]){P[i] = cur;if(!DFS(i))break;}return false;//上面一个break,下面false接住
}bool ifConnectToSource(int cur){if(cur == source)return true;if(cur == -1)return false;return ifConnectToSource(P[cur]);
}int main(){IN = fopen("my_input.txt", "r"); OUT = fopen("my_output.txt", "w");//fscanf(IN,"%d %d %d", &v_num,&e_num,&source);first = 1; last = first + v_num - 1;memset(G,0,sizeof(G));while(e_num--){int v1,v2; fscanf(IN,"%d %d", &v1,&v2);G[v1][v2] = G[v2][v1] = 1;}memset(P,-1,sizeof(P));P[source] = -2;energy = 4;DFS(source); fprintf(OUT,"\n");for(int i = first; i <= last; i++)fprintf(OUT,"%d\t%d\n",i,P[i]);//fclose(IN); fclose(OUT);return 0;
}

这里用P[i]==-1来判断是否访问过的方式有一个问题，
那就是P[source]是谁。。。
按理说source没有前驱，
那么P[source] = -1，但是等于-1又会被判定是没有访问过的节点
。。。。。。。。。。。。所以。。。。所以。。。把P[source]初始化为-2。
BFS用P[i]判断是否访问过的时候不用考虑这一点，因为BFS都是用extract的。。。

另外一点就是，
如果能量耗完，我应该全盘退出，不只是退出这个递归，而是整个所有进行中的DFS。
因此我必须把DFS改成bool类型，以传递某种信息。
可以用goto，但是肯定不能用goto。

输出

遇到某一条边就全盘退出

假设小明从源点开始DFS，图上某一条边是陷阱，问题是小明不知道那条边是陷阱，结果小明被困在了陷阱里。
请你复盘小明走过的路径。
和上一题一样，需要合理安排路径，使得可以实现全盘退出。

//DFS_TRAP.c
# include<stdio.h>
# include<stdlib.h>
# include<stdbool.h>
# include<string.h>
FILE *IN,*OUT;# define N 9int v_num,e_num,first,last,G[N][N],source;
int P[N];
int trap_from,trap_end;bool DFS(int cur){fprintf(OUT,"%d ",cur);for(int i = first; i <= last; i++)if(P[i]==-1 && G[cur][i]){P[i] = cur;if(cur == trap_from && i==trap_end)return false;//合理设计代码实现全盘退出if(!DFS(i))return false;//如果下面的dfs传来false，全盘退出}return true;//上面全部return false拦截掉,如果能走到下面这里true代表小明不会遇到坑
}bool ifConnectToSource(int cur){if(cur == source)return true;if(cur == -1)return false;return ifConnectToSource(P[cur]);
}int main(){IN = fopen("my_input.txt", "r"); OUT = fopen("my_output.txt", "w");//fscanf(IN,"%d %d %d", &v_num,&e_num,&source);first = 1; last = first + v_num - 1;memset(G,0,sizeof(G));while(e_num--){int v1,v2; fscanf(IN,"%d %d", &v1,&v2);G[v1][v2] = G[v2][v1] = 1;}memset(P,-1,sizeof(P));P[source] = -2;trap_from = 5;trap_end = 7;//5和8之间不存在边，把这个改成8试试看小明会不会摔死//我试过了，小明不会摔死DFS(source); fprintf(OUT,"\n");for(int i = first; i <= last; i++)fprintf(OUT,"%d\t%d\n",i,P[i]);//fclose(IN); fclose(OUT);return 0;
}

输出

利用DFS进行TOPO排序

DFS模板的最下面一行可以写"blacken this",这代表这个节点已经全盘结束了。
当根节点全盘结束的时候整个DFS就都结束了。
在这里我们每次finish一个节点，就push_front到一个队列里。这个队列的结果就是拓扑排序。
在拓扑排序里面，任意取出一个有序对，左边的总是能流向右边的。
我这里也提供了DFS的输出结果，以供对比。第二行是topo排序。
注意是有向图。

//DFS_TOPO.c
//拓扑排序是给有向图用的，所以我把ReadData里面给改成有向图了
# include <stdio.h>
# include <stdlib.h>
# include <stdbool.h>
# include <string.h>
FILE *IN,*OUT;# define N 9int n,m,first,last,source,G[N][N],color[N],topo[N],Q[N],front,rear;void DFS(int cur){color[cur] = 1;fprintf(OUT,"%d ",cur);for(int i = first; i <= last; i++)if(!color[i] && G[cur][i]) DFS(i);//topo.pushfront(cur);Q[front] = cur;front = (front+N-1)%N;
}void ReadData(){fscanf(IN,"%d%d%d",&n,&m,&source);first = 1; last = first + n - 1;memset(G,0,sizeof(G));while(m--){int v1,v2;fscanf(IN,"%d%d", &v1,&v2);G[v1][v2] = 1;}
}int main(){IN = fopen("my_input.txt","r"); OUT = fopen("my_output.txt","w");ReadData();memset(color,0,sizeof(color));memset(topo,0,sizeof(topo));front = rear = 0;//实际上只要是[0,N)的任何一个值都可以，只要一样就得了DFS(source);fprintf(OUT,"\n");while(front != rear){//extract frontfront = (front+1)%N;int tmp = Q[front];fprintf(OUT,"%d ",tmp);}//Awesomefprintf(OUT,"\n");fclose(IN); fclose(OUT);return 0;
}

输出

1 2 3 4 5 7
1 2 3 4 7 5

利用DFS记录结束时间

同理，只要一个全局变量即可实现。

//DFS_TIME.c
# include<stdio.h>
# include<stdlib.h>
# include<stdbool.h>
# include<string.h>
FILE *IN,*OUT;# define N 9int v_num,e_num,first,last,G[N][N],source,color[N];
int time,finish_time[N];void DFS(int cur){++time;color[cur] = 1; fprintf(OUT,"%d ",cur);for(int i = first; i <= last; i++)if(!color[i] && G[cur][i])DFS(i);finish_time[cur] = ++time;
}int main(){IN = fopen("my_input.txt", "r"); OUT = fopen("my_output.txt", "w");//fscanf(IN,"%d %d %d", &v_num,&e_num,&source);first = 1; last = first + v_num - 1;memset(G,0,sizeof(G));while(e_num--){int v1,v2; fscanf(IN,"%d %d", &v1,&v2);G[v1][v2] = G[v2][v1] = 1;}memset(color,0,sizeof(color));memset(finish_time,0,sizeof(finish_time));time = -1;DFS(source); fprintf(OUT,"\n");for(int i = first; i <= last; i++)fprintf(OUT,"%d\t%d\n",i,finish_time[i]);//fclose(IN); fclose(OUT);return 0;
}

输出

1 2 3 4 5 7 6
1   13
2   12
3   11
4   10
5   7
6   9
7   6
8   0

结束语

DFS看起来比BFS代码少，但实际上出处都是坑啊！！！递归就是那么优美而紧凑的东西。