POJ 1270 following Orders (topological sort +dfs)

Approximate test instructions: Each sample contains two lines, the first line is entered n characters, and may be unordered. The second line is entered in pairs of a B, representing a to precede B. All of the outputs conform to such sequences.

Idea: A very obvious sort of topology. To output all of the sequences, DFS is performed from a point with a degree of 0, each time selecting a point with an entry level of 0, adding an output sequence and reducing the degree of the point adjacent to it. After Dfs is over, change the status back.

#include <stdio.h> #include <algorithm> #include <set> #include <map> #include <vector># Include <math.h> #include <string.h> #include <stack> #include <queue> #define LL long long# Define _ll __int64using namespace Std;const int maxn = 110;char contents[30];bool mapp[60][60];int n = 0;char t,t1,t2;int            Flag;int in[30];int vis[30];int ans[30];void dfs (int num) {if (num = = N) {for (int i = 0; i < n; i++)        printf ("%c", Ans[i]);        printf ("\ n");    Return } for (int i = 0; i < n; i++) {if (!vis[i] &&!in[contents[i]-' A "]) {for (int j = 0; J < N; J + +) {if (mapp[contents[i]-' a '][contents[j]-' a ']) {in[Conten                ts[j]-' a ']--;            }} Ans[num] = Contents[i];                        Vis[i] = 1;                        DFS (NUM+1); Note Change the status back to for (int j = 0; J < N J + +) {if (mapp[contents[i]-' a '][contents[j]-' a ']) {in[Conten                ts[j]-' a ']++;        }} Vis[i] = 0;    }}}int Main () {flag = 0;            while (~SCANF ("%c", &t)) {if (T >= ' a ' && t <= ' z ') {contents[n++] = t;        Continue        } else if (t = = ") continue;            else {if (flag) printf ("\ n");            flag = 1; Sort (contents,contents+n);            Because the input may be unordered, sort memset (in,0,sizeof (in)) first;            memset (mapp,false,sizeof (MAPP));            memset (vis,0,sizeof (VIS));                while (~scanf ("%c%c", &t1,&t2)) {mapp[t1-' a '][t2-' a '] = 1;                in[t2-' a ']++;            if (getchar () = = ' \ n ') break;            } dfs (0);        n = 0; }} return 0;}