【题目大意】:
给出 A“<”B 的关系图
①在第i个关系之chu后,得出XXXX
②在第i个关系之后 出现错误(环)
③全部关系之后仍然无法排序
【解题思路】:1.读一个数判断是否是②情况。
2.利用Floyd更新 任意两元素之间关系,当更新数达到(n*n-n)/2 之后,得到①情况。
3.数据全部读入之后,仍没有答案,得到③情况。
1 #include "cstdio" 2 #include "iostream" 3 #include "cstring" 4 using namespace std; 5 int E[30][30]; 6 int sum[30]; 7 void putout(int n) 8 { 9 int s; 10 for (int i = 1; i <= n; i++) 11 { 12 s = 1; 13 for (int j = 1; j <= n; j++) 14 if (E[i][j] && i != j) s++; 15 sum[s] = i; 16 } 17 for (int i = n; i >= 1 ; i--) 18 { 19 printf("%c", sum[i] + 'A' - 1); 20 } 21 22 } 23 int main() 24 { 25 int n , m; 26 bool wait; 27 int ans = 0; 28 while (cin >> n >> m && (n + m)) 29 { 30 memset(E, 0, sizeof(E)); 31 char a, b; 32 wait = true; 33 ans = (n * n - n) / 2; 34 for (int i = 1; i <= m ; i++) 35 { 36 scanf(" %c"<"%c", &a, &b); 37 38 if (wait) 39 { 40 if (E[b - 'A' + 1][a - 'A' + 1] == 1 || (a - 'A' + 1) == (b - 'A' + 1)) 41 { 42 wait = false; 43 printf("Inconsistency found after %d relations. ", i); 44 continue; 45 } 46 if (E[a - 'A' + 1][b - 'A' + 1] == 0) 47 { 48 E[a - 'A' + 1][b - 'A' + 1] = 1; 49 ans--; 50 } 51 for (int k = 1 ; k <= n ; k++) 52 for (int j = 1; j <= n; j++) 53 for (int z = 1; z <= n; z++) 54 if (E[j][z] == 0 && E[j][k] == 1 && E[k][z] == 1 && (j != z)) 55 { 56 ans--; 57 E[j][z] = 1; 58 } 59 if (!ans) 60 { 61 wait = false; 62 printf("Sorted sequence determined after %d relations: ", i); 63 putout(n); 64 printf(". "); 65 continue; 66 } 67 } 68 } 69 if (wait) printf("Sorted sequence cannot be determined. "); 70 } 71 }