75 lines
2.3 KiB
C++
75 lines
2.3 KiB
C++
#include <bits/stdc++.h>
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using namespace std;
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const int N = 26;
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int n, m;
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bool g[N][N];
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bool st[N];
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int check() {
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for (int i = 0; i < n; i++)
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if (g[i][i]) return 2; // 矛盾
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for (int i = 0; i < n; i++)
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for (int j = 0; j < i; j++)
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if (!g[i][j] && !g[j][i]) // 待继续
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return 0;
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return 1; // 找到顺序
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}
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string getorder() { // 升序输出所有变量
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char s[26];
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for (int i = 0; i < n; i++) {
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int cnt = 0;
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// f[i][j] = 1表示i可以到达j (i< j)
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for (int j = 0; j < n; j++) cnt += g[i][j]; // 比i大的有多少个
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// 举个栗子：i=0,表示字符A
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// 比如比i大的有5个，共6个字符：ABCDEF
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// n - cnt - 1 = 6-5-1 = 0,也就是A放在第一个输出的位置上, 之所以再-1，是因为下标从0开始
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s[n - cnt - 1] = i + 'A';
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}
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// 转s字符数组为字符串
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string res;
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for (int i = 0; i < n; i++) res = res + s[i];
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return res;
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}
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int main() {
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while (cin >> n >> m, n || m) {
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memset(g, 0, sizeof g);
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int type = 0, t;
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for (int i = 1; i <= m; i++) {
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char str[5];
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cin >> str;
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int a = str[0] - 'A', b = str[2] - 'A';
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// a<b,那么，不需要完全的重新计算完整的传递闭包，只需要把与a,b相关的变更进行记录大小关系即可
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if (!type) {
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g[a][b] = 1;
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for (int x = 0; x < n; x++) {
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if (g[x][a]) g[x][b] = 1; // 所有比a小的x,一定比b小
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if (g[b][x]) g[a][x] = 1; // 所有比b大的x,一定比a大
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for (int y = 0; y < n; y++)
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if (g[x][a] && g[b][y])
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g[x][y] = 1;
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}
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type = check();
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if (type) t = i;
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}
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}
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if (!type)
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puts("Sorted sequence cannot be determined.");
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else if (type == 2)
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printf("Inconsistency found after %d relations.\n", t);
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else {
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string ans = getorder(); // 输出升序排列的所有变量
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printf("Sorted sequence determined after %d relations: %s.\n", t, ans.c_str());
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}
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}
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return 0;
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} |