【POJ1094】Sorting it all out

题目链接：https://www.acwing.com/problem/content/345/

题目大意：给定 (m) 个不等式关系 , 确定变量间的大小关系 , 或找出不等式关系间的矛盾

solution

如果(i < j) , (j < k) , 则 (i < k) ,因此这些不等式关系间具有传递性 , 所以这是一道经典的传递闭包问题 ,可以用(Floyd) 传递关系

令 (g[i][j] = 1) 表示 (i < j) , (g[i][j] = 0) 表示 (i geq j) 或不确定 , 其中(g[i][j]) |= (g[i][k]) & (g[k][j]) , 最后依次遍历 , 如果 (g[i][j] = 1) 且 (g[j][i] = 1) , 则发生矛盾 , 如果 (g[i][j] = 0) 且 (g[j][i] = 0) , 则无法确定 , 否则 , 可以确定 (i) 与 (j) 之间的唯一关系 , 输出时每次找到最小的输出并排除即可

时间复杂度: (O(mn^2))

code

#include<bits/stdc++.h>
using namespace std;
template <typename T> inline void read(T &FF) {
	int RR = 1; FF = 0; char CH = getchar();
	for(; !isdigit(CH); CH = getchar()) if(CH == '-') RR = -RR;
	for(; isdigit(CH); CH = getchar()) FF = FF * 10 + CH - 48;
	FF *= RR;
}
inline void file(string str) {
	freopen((str + ".in").c_str(), "r", stdin);
	freopen((str + ".out").c_str(), "w", stdout);
}
const int N = 1e3 + 10;
char ai, bi, ci;
int pd[N][2], g[30][30], n, m;
int check(int xi) {
	memset(g, 0, sizeof(g));
	for(int i = 1; i <= xi; i++)
		g[pd[i][0]][pd[i][1]] = true;
	for(int k = 1; k <= n; k++)
		for(int i = 1; i <= n; i++)
			for(int j = 1; j <= n; j++)
				g[i][j] |= g[i][k] & g[k][j];
	int flag = 0;
	for(int i = 1; i <= n; i++)	
		for(int j = 1; j <= n; j++)
			if(i != j) {
				if(g[i][j] && g[j][i]) return false;
				else if(!g[i][j] && !g[j][i]) flag = true;
			} 
			else if(g[i][j] == 1) return false;
	return flag + 1;
}
void print(int xi) {
	printf("Sorted sequence determined after %d relations: ", xi);
	int out[30] = {0};
	for(int i = 1; i <= n; i++) {
		for(int j = 1; j <= n; j++)
			if(!out[j]) {
				int flag = 0;
				for(int k = 1; k <= n; k++)
					if(!out[k] && g[k][j]) {
						flag = 1;
						break;
					}
				if(!flag) {
					cout << char(j + 'A' - 1), out[j] = 1;
					break;
				}
			}
	}
	cout << "." << endl;
}
int main() {
	//file("");
	while(true) {
		read(n), read(m);
		if(!n && !m) break;
		for(int i = 1; i <= m; i++) {
			cin >> ai >> bi >> ci;
			if(bi == '<') pd[i][0] = ai - 'A' + 1, pd[i][1] = ci - 'A' + 1; 
			else pd[i][1] = ai - 'A' + 1, pd[i][0] = ci - 'A' + 1;
		}
		int fg = 0;
		for(int i = 1; i <= m; i++)
			if(!check(i)) {
				printf("Inconsistency found after %d relations.
", i);
				fg = 1; break;
			}
			else if(check(i) == 1) {
				print(i);fg = 1; break;
			}
		if(fg)  continue;
		puts("Sorted sequence cannot be determined.");
	}
	return 0;
}