POJ1094 Sorting It All Out（拓扑排序 or Floyd）

曾经终败给现在 2021-11-16 07:56 219阅读 0赞

## 题意：对于N个大写字母，给定它们的一些偏序关系，要求判断出经过多少个偏序关系之后可以确定它们的排序或者存在冲突，或者所有的偏序关系用上之后依旧无法确定唯一的排序。 ##

## 分析：拓扑排序模板题，发现也可以用最短路Floyd算法解决，详见《算法竞赛进阶指南》P360。 ##

## 代码： ##

#include <cstdio>
    #include <cstring>
    #include <iostream>
    #include <algorithm>
    using namespace std;
    const int N = 30;
    int n, m, d[N][N], e[N][N];
    
    int floyd() {
    	memcpy(e, d, sizeof(e));
    	for (int k = 0; k < n; k++)
    		for (int i = 0; i < n; i++)
    			for (int j = 0; j < n; j++) {
    				e[i][j] |= e[i][k] & e[k][j];
    				if (e[i][j] == e[j][i] && e[i][j] && i != j) return -1;
    			}
    	for (int i = 0; i < n; i++)
    		for (int j = 0; j < n; j++)
    			if (e[i][j] == e[j][i] && !e[i][j] && i != j) return 0;
    	return 1;
    }
    
    void Sorting_It_All_Out() {
    	memset(d, 0, sizeof(d));
    	bool flag = 1;
    	for (int i = 1; i <= m; i++) {
    		char s[6];
    		scanf("%s", s);
    		d[s[0]-'A'][s[2]-'A'] = 1;
    		if (flag) {
    			int now = floyd();
    			if (now == -1) {
    				printf("Inconsistency found after %d relations.\n", i);
    				flag = 0;
    			} else if (now == 1) {
    				printf("Sorted sequence determined after %d relations: ", i);
    				pair<int, char> ans[N];
    				for (int j = 0; j < n; j++) {
    					ans[j].first = 0;
    					ans[j].second = 'A' + j;
    				}
    				for (int j = 0; j < n; j++)
    					for (int k = 0; k < n; k++)
    						if (e[j][k]) ++ans[j].first;
    				sort(ans, ans + n);
    				for (int j = n - 1; j >= 0; j--) printf("%c", ans[j].second);
    				puts(".");
    				flag = 0;
    			}
    		}
    	}
    	if (flag) puts("Sorted sequence cannot be determined.");
    }
    
    int main() {
    	while (cin >> n >> m && n) Sorting_It_All_Out();
    	return 0;
    }