解决Spoj平台Paradox问题的简单算法-优快云博客

本文介绍了一种解决Spoj平台Paradox问题的高效算法，通过检查路径长度来判断是否存在矛盾循环，从而识别出导致悖论的情况。

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http://www.spoj.com/problems/PARADOX/

这题不知道为啥被归成dfs，一艘题解，甚是惊讶，竟然找出规律来了，佩服+1

多种讨论http://www.quora.com/What-is-the-algorithmic-approach-to-solving-the-problem-Paradox-on-SPOJ

A straightforward problem, if you can observed that there will be PARADOX if and only if there will be cycle containing an statement $i$ and having odd number of false implies in that cycle.
For example,
A -> B $false$
B-> C $false$
C -> A $false$
There will be a contradiction at vertex A as in the Cycle A->B->C->A there will be $odd$ number of false.
A simple ASAP algorithm will check that. Simply assign edge weight 1 if statement a is saying statement b is false, and 2 as true and check whether path length from A to A having odd length.

#include <iostream>
using namespace std;

#define MX 10000000

int main() {
	// your code goes here
	int T;
	while(1)
	{
		int N,i,j,k;
		cin>>N;	
		if(N==0)
			return 0;
		int dis[N+1][N+1];
		for(i=0; i<=N ; i++)
			for(j=0 ; j<=N ; j++)
				dis[i][j] = MX;

		for(i=1 ; i<=N ; i++)
		{
			int x;
			string str;
			cin>>x>>str;
			if(str == "true")
				dis[i][x] = 2;
			else
				dis[i][x] = 1;
		}

		for(k=1;k<=N;k++)
		{
			for(i=1;i<=N;i++)
			{	
				for(j=1;j<=N;j++)
				{
					if(dis[i][j]>dis[i][k]+dis[k][j])
					{
						dis[i][j]=dis[i][k]+dis[k][j];
					}
				}
			}
		}

		int temp=0;
		for(i=1 ; i<=N ; i++)
		{
			if( dis[i][i] < MX/ 2 && dis[i][i] % 2 == 1)
			{
				cout<<"PARADOX"<<endl;
				temp = 1;
				break;
			}

		}
		if(temp == 0)
		{
			cout<<"NOT PARADOX"<<endl;
		}
	}
	return 0;
}